INTRODUCTION SIMPLE HARMONIC MOTION LINEAR SIMPLE ...

COLLEGES: ANDHERI / BORIVALI / CHEMBUR / DADAR / KALYAN / KHARGHAR / NERUL / POWAI / THANE 74

INTRODUCTION A motion which repeats itself periodically in equal time intervals is said to be periodic or harmonic. Repeating itself implies the movement through the same point with the same velocity again and again. We come across several examples of periodic motion in day to day life. The revolution of earth round the sun which repeats itself in one year, the revolution of moon round the earth, which repeats itself in one lunar month, the movement of the bob of a simple pendulum, the movement of the crank fitted in a vehicle moving with constant velocity etc. A motion involving to and fro or back and forth motion, is said to be oscillatory. In such a motion, the body crosses a point with two different velocities equal in magnitude but opposite in directions alternately. In other words, the body keeps on retracing the path, between two fixed points. Note that all undamped oscillatory motions, which we come across in daily life is necessarily periodic. However, all periodic motions need not be necessarily oscillatory. Consider, for example, the motion of the bob of a clock pendulum. Since it moves to and fro, between two fixed points, its motion is oscillatory. Moreover, it keeps on repeating itself in a fixed interval (usually two seconds and hence it is periodic too). Now, let us consider the motion of earth around the sun. We have seen earlier that it is, no doubt, periodic, but the motion always takes along the same way and earth never “retraces” its path. Hence, it is not an oscillatory motion.

SIMPLE HARMONIC MOTION

It is a special case of oscillatory motion, in which the acceleration of the vibrating particle (or body), at any position, varies directly as its displacement from a fixed point, (which may or may not lie along the line of motion) and is always directed towards that fixed point. Thus, simple harmonic motion (abbreviated as SHM) is a case of variable acceleration; however, the variation takes place in a regular and periodic fashion.

LINEAR SIMPLE HARMONIC MOTION

In this case, the motion of the particle (or body) involved is to and fro along a straight line, besides the other necessary conditions. Let a particle P perform oscillatory motion between two fixed points A and B. Let O be the mid point of A and B. (fig.) If the particle P oscillates about point O in such a way that its acceleration ‘f, at any position when its displacement from O is x, can be mathematically expressed as f x and is directed towards O, then its motion will be SHM. Here, the point ‘O’ is known as mean or stable or equilibrium

BPO

Ax

or neutral position, and in case of “simple” harmonic motion, the maximum displacement, of the particle on either side of the mean position is the same i.e. OA = OB. Now, taking the motion of the particle to be along X-axis, SHM can be mathematically expressed as

f x or 2f x

…….. (1)

Here, the negative sign stands for the fact that the direction of acceleration is opposite to that of displacement, and 2 is a positive constant. Again, if m be the mass of the oscillating particle (or body), then, multiplying both sides of eqn. (1) by m, we have

x)m(m 2 f or x)k(F

........ (2)

where Fm

f is the force acting on the particle and k = m2 a positive constant.

Equations (1) or (2) can equally be used to show that a given motion is SHM. From equation (2) it is evident that x = 0 ; F = 0 which shows that the particle experiences no force, when at mean position, it will not oscillate. However, if it is disturbed even slightly by external force, then new forces (restoring forces) should be set up in the system which should tend to bring the particle to its mean position. Depending upon the nature of restoring forces, we have several types of SHM. The restoring forces can be electrical, gravitational, magnetic, elastic etc.


Velocity and displacement of a particle executing SHM. From equation (1),

f = 2x [Omitting the vector sign]

or dt

dv = 2x

dt

dx.

dx

dv = 2x vdv = 2x dx

Integrating both sides, we get 1

222

C2

x

2

v

where C1 is a constant of integration. Since the velocity of the particle is zero at its extreme position x = a (say)

0 = 1

22

C2

a

C1 =

2

a 22

Substituting for C1, we get

2

a

2

x

2

v 22222

or v2 = 2(a2 x2)

or v = + (a2 x2)1/2 ........ (3) Again,

v = dt

dx

From equation (3). dtxa

dx22

Integrating both sides, we get

sin1

a

x = t + C2

Where C2 is a constant of integration; let at t = 0 ; x = x0

Then, sin1

a

x 0 = C2 = (say)

Then the above equation can be rewritten as

sin1

a

x = t +

a

x = sin (t + )

x = a sin (t + ) …….. (4) Rewriting equation (3), by substituting for ‘x’ in terms of t from (4), we have v = [a2 a2 sin2 (t + )]1/2 v = a cos (t + ) …….. (5) Also, rewriting equation (1), by substituting for x in terms of t from (4), we have f = 2 a sin (t + ) …….. (6) Equations (1) and (3), express the acceleration and velocity of a particle executing SHM in terms of displacement. Equation (4), (5) and (6) express displacement, velocity and acceleration, in terms of time. Illustration 1: A particle executes SHM between two points separated by 10 cm. If the force experienced by it be

2N at a displacement of 1 cm from the mean position, then find the maximum force experienced by it.

Sol. Evidently, the maximum displacement = 2

10 = 5 cm.

For a particle executing SHM the force experienced is given by F = kx …….. (1) and Fmax = k(xmax) …….. (2)


B

O P

Acm5x

cm10

Dividing (2) by (1)

x

x

F

F maxmax 1

5

2

Fmax Fmax = 10N.

Illustration 2: A particle executing SHM oscillates between two fixed points separated by 20 cm. If its maximum

velocity be 30 cm/s, find its velocity when its displacement is 5 cm from its mean position. Sol. Evidently, the max. Displacement

= 2

20 = 10 cm

For a particle executing SHM the speed at a displacement ‘x’ is given by

v = 22 xa …….. (1) and vmax = a …….. (2) Dividing (1) by (2), we get

22

max

xa

a

v

v

3

32

35

10

x10

10

v

v22

max

v = 30

3

32 = 315 cm/s.

TERMS CONCERNED WITH SHM

(a) Amplitude: It is the maximum displacement of the particle executing SHM from its mean position. From equation (4), it is obvious that the displacement varies between the limits a to +a as sin(t + ) varies continuously between the limits 1 and +1 respectively. Thus, amplitude = |xmax| = a (b) Time Period: or [Period of Oscillation]. We have discussed earlier that all SHMs are periodic motions which repeat themselves in equal time intervals. This minimum time interval is known as time period for the oscillations. Let T be the time period for the oscillation, then from equation (4), x = a sin (t + ) = a sin [ (t + T) + ] sin(t + ) = sin (t + + T) Since all sine and cosine functions vary periodically with an angle 2

T = 2 T = 2

(c) Angular Frequency: The number of revolutions (expressed in radian) performed per unit time is known as angular frequency. (Each oscillation corresponds to one revolution; see more in the next article) In a time T (time period), no. of revolution covered = 1

In a time of 1 sec. no. of revolutions covered = T

1

Angle described in each revolution = 2

Angle described in T

1 revolution =

T

2

Thus, angular frequency = T

2 =

The number of oscillations described per unit time is known as the frequency of oscillations.


Evidently, frequency n = T

1

(d) Phase: We have seen earlier, that the displacement, velocity and acceleration of a particle executing SHM vary periodically with the angle (t +) associated with the sine or cosine term. Knowing this angle, we can be sure of its position as well as state of motion as to how and where is it oscillating. This angle (t + ) is known as the “phase” of the oscillating particle. Since the phase is a time dependent factor, it will be more worthwhile to speak in terms of instantaneous phase (i.e., phase at any instant). Taking the displacement equation as x = a sin (t +) we have at t + = 0 ; x = 0 at t + = /2 ; x = a at t + = ; x = 0 at t + = 3/2 ; x = a and so on. Notice that, as time varies indefinitely, phase keeps on changing, between 0 to 2; the reason being (t + ) = (t + ) + 2n where n I.

(e) Phase constant [Epoch] : The phase of a particle executing SHM, initially (i.e., at the instant when time was reckoned) is known as the initial phase or phase constant or epoch.

Since instantaneous phase = t + , so Initial phase can be obtained by putting t = 0 Phase constant = . As a special case, if at the instant of start of motion phase be zero, then = 0. All, the equation can be rewritten by replacing (t + ) with t. Considering the displacement equation as x = a sin (t + )

Let us put =

'2

the displacement equation becomes

x = a sin

'

2t

= a cos (t + ’) Differentiating w.r.t. time t v = a sin (t + ’) and f = a 2 cos (t + ’) = 2 x. The above equation too represent various parameters associated with SHM. Relation between Phase Difference and Time Difference. Let the phase of a particle at time t1 be 1, then 1 = t1 + …….. (a) Let the phase of the particle change to 2, at time t2, then 2 = t2 + …….. (b) Subtracting (a) from (b) Phase différence 2 1 = (t2 t1) or = t

Putting t = T and = T

2, we have = 2

Thus, a phase difference of 2 is equivalent to a time difference of T. Similarly, a phase’ difference of is equivalent to a time difference of T/2, and so on.


GRAPHS CONCERNING SHM (a) Graph of Displacement Versus Time Assuming the displacement equation to be x = a sin t, clearly the graph will be a sine curve. (Figure) Putting x = 0, we have sin (t) = 0

tn = n, where n I tn = n

Putting n = 0, 1, 2, 3 etc. we get

t0 = 0, t1 =

, t2 = 2

, t3 = 3

etc.

and putting x = + a (i.e., maximum displacement) we get sin (t) = + 1

tn = (2n + 1)2

where n I

Putting n = 0, 1, 2 etc. we get

t0 =

2, t1 =

2

3, t2 =

2

5 etc.

(b) Graph of Velocity Versus Time

Assuming x = a sin t, we get on differentiating dt

dx = v = a cos t

Putting v = 0; we get cos t = 0

tn = (2n + 1)2

; where n I

Putting n = 1, 2, etc.

t0 =

2, t1 =

2

3, t2 =

2

5 etc.

Putting v = + a (velocity amplitude), we get cos wt = + 1 tn = n where n I

tn = n

, putting n = 0, 1, 2 etc.

we get t0 = 0 ; t1 =

, t2 = 2

etc.

The graph is a cosine curve (Figure) (c) Graph of Acceleration Versus Time Assuming x = a sin t, we have f = 2 x = 2 a sin t Putting f = 0, the instants at which f = 0, can be found: sin t = 0 tn = n

Putting n = 0, 1, 2, …… t0 = 0, t1 =

, t2 = 2

etc.

Putting f = 2 a; we have sin t = + 1 (acceleration amplitude)

tn = + (2n + 1)2

Putting n = 0, 1, 2, etc.

t0 =

2, t1 =

2

3, t2 =

2

5 etc.

Clearly the graph is a sine curve (Figure)

)x(disp

a

a

2

2

3

2

2 time)t(

O

)v(vel

a

a

2

2

3

2

4 time)t(

O

)f(.acc

a2

a2

2

2

3

2

4 time)t(

O


(d) Graph of Acceleration Versus Displacement Since acceleration (f) = 2 (x). The variation is linear (Figure) (e) Graph of Velocity Versus Displacement We have seen earlier that

v = 22 xa v2 = 2 (a2 x2)

2

2v

= a2 x2

22

2

2

2

a

v

a

x

= 1

The graph is an ellipse. (Figure) (f) Graph of Variation between Displacement (x) and Phase (t + ). From the equation x = a sin (t + ) x = 0 at (t + ) = n where n I

and x = + a at (t + ) = (2n + 1)2

, where n I

The graph is a sine curve (Figure) (g) Graph of Velocity Versus Phase From the equation v = a cos (t + )

v = 0 at (t + ) = (2n + 1)2

, where n I and

v = + a at (t + ) = n , where n I The graph is shown in Figure Illustration 3: A particle executes SHM with an angular frequency = 4 rad/sec. If it is at its extreme position

initially, then find the instants, when it is at a distance 2

3 times its amplitude from the mean

position. Sol. Let a be the amplitude and the phase constant for the oscillation. The displacement equation can be written as x = a sin (4t + ) [ = 4 rad/sec] Since at t = 0 ; x = + a

4 (0) + = 2

=

2

x = a sin (4t + /2) = a cos (4t)

Now let tn be the instant, when the particle’s displacement is 2

a3, nth time (where n I).

Evidently + 2

a3 = a cos (4tn)

4tn = n + 6

tn =

24

1

4

n

Taking only the positive values of ‘t’

t1 = 24

1s, t2 =

24

5s, t3 =

24

7s, t4 =

24

11s, t5 =

24

13s, ect.

Illustration 4: A particle execute SHM with an amplitude 8 cm and a frequency 10 sec1. Assuming the particle to

be at a displacement 4 cm, initially moving in the positive direction, determine its displacement equation and the maximum velocity and acceleration.

Sol. Given a = 8 cm and = 2n

f

x

)v(vel

)x(disp

a

a

aa

)x(

a

a

2

2

3 2Phase

)t(

O

2

5

)v(

a

a

2

2

3 2 Phase)t(

O

2

5

velocity

3


= 20 rad/sec Let the phase constant be . The displacement equation can be written as x = 8 sin (20t + ) Given, at t = 0 ; x = 4 cm 4 = 8 sin (20(0) + )

sin () = 2

1 =

6

The displacement equation becomes

x = 8 sin

6

t20

Differentiating the above equation w.r.t. time ‘t’

dt

dx = v = 160 cos

6

t20

vmax = + 160 cm/s [when cos (20t + /6) = + 1] Differentiating once again w.r.t. time t.

2

2

dt

xd = f = 3200 2 sin (20t + /6)

fmax = 3200 2 cm/s [when sin (20t + /6) = + 1]

Illustration 5: The equation of particle executing simple harmonic motion is x = (5 m) sin 1( s )t3

. Write

down the amplitude, time period and maximum speed. Also find the velocity at t = 1 s. Sol. Comparing with equation x = A sin (t + ), we see that the amplitude = 5 m,

and time period 1

2 22s

s

The maximum speed = A = 5 m × s–1 = 5 m/s.

The velocity at time t = dx

dt = A cos (t + )

At t = 1 s, v = (5 m) ( s–1) cos 5

= –

5

2

m/s.

Illustration 6: If the maximum speed and acceleration of a particle executing SHM be 20 cm/s and 100 cm/s2, find the time period of oscillation.

Sol. Given |vmax| = a = 20 cm/s ……… (1) and |fmax| = a2 = 100 cm/s2 ……… (2) Dividing equation (2) by (1) we get = 5 rad/sec.

Time period T =

5

22 = 0.4 sec.

Illustration 7: If a particle executing SHM possesses a speed of 10 cm/s, when at mean position, find its speed when at a displacement of half the amplitude.

Sol. Let ‘a’ be the amplitude of oscillation

For velocity v = 22 xa Squaring, we get v2 = 2 (a2 x2) Given at x = 0 ; v = 10 100 = 2a2 …….. (1)

and let at x = 2

a; v = v’


B O A

2

ax

cm10)positionmean(

(v')2 = 2

4

a3 2

…….. (2)

Dividing equation (2) by (1), we get

4

3

10

'v2

v’ =

2

3 10 = 53 cm/s

Illustration 8: Two particle execute SHM with same frequency and amplitude along the same straight line. They

cross each other, at a point midway between the mean and the extreme position. Find the phase difference between them

Sol. Let ‘a’ be the amplitude and 1 and 2 their respective phases at any instant, when they cross each other. Then their displacement equation can be written as x = a sin (phase) For the two particle we have

21 sina2

aandsina

2

a taking the least value, 1 =

6

and taking the next possible value (∵ 1 ≠ 2 )

2 = 6

5

Moreover, from V

= a cos (phase)

If 1v and 2v be the respective velocities of the particles, then 1v = 2v i.e., a cos (1) = a cos (2) cos 2 = cos ( 1)

If 1 = ,6

then 2 =

6

7or

6

5

required phase difference = 2 1 3

2 rad.

NOTE: The phase difference can be in general 3

2 + 2n ; n I+.

Illustration 9: A particle executes SHM with time period of 2 sec; Find the time taken by it to move from one extreme position to the nearest mid point between mean position and extreme position.

Sol. Let ‘a’ be the amplitude of oscillation then evidently = T

2 = rad/sec.

The displacement equation can be written as x = a sin(t + ) Let, for simplicity, the time be reckoned from the instant when the particle was at its extreme position. Thus, at t = 0; x = a

a = a sin () = 2

Thus, x = a sin

2

t

Let at t = t1 (+ve) x = 2

a

2

a = a sin

2

t1


2

t1 = 3

t6

51

t1 = 3

1 sec.

Illustration 10: A particle is executing SHM with amplitude ‘a’ and maximum velocity v0. Find the displacement

when its velocity is 03v.

2

Sol. From the equation for speed v2 = 2 (a2 x2) 2

0v = 2 a2 ….. (1)

and 4

v3 20 = (a2 x2) 2 ….. (2)

Dividing (1) by (2), we get

2

2

22

2

a

x1

4

3or

xa

a

3

4

4

1

a

x2

2

x = 2

a

MOTION OF A FOOT OF PERPENDICULAR UPON ANY DIAMETER OF A

PARTICLE EXECUTING UNIFORM CIRCULAR MOTION Consider a particle (reference particle) moving along a circle of radius a, with uniform angular speed . The circle is known as circle of reference. Let X’X and Y’Y be any two mutually perpendicular diameters, which can be taken as X and Y axes respectively. Let M be the foot of perpendicular drawn from the Particle P upon the diameter X’X. As will be seen ahead that, as the reference particle executes uniform circular motion along the circle, the foot of perpendicular M (or the projection of P upon the diameter X’X) executes SHM along the diameter X’X. Let P0 be the initial position of the reference particle and corresponding angle made by the radius vector joining center o to particle P0 with the axis OY’ be . Further, let in a time t, the radius vector OP0 rotate by an angle = t, and reach the new position OP. M is the projection of P along X’X. The displacement of M (referred to ‘O’ the center of circle of reference) will be given by OM = x = a sin (t + ) [From rt. angled MOP] ….. (a) The velocity of the projection M, will be the component of linear velocity of particle p along X axis. v = a cos (t + ) ….. (b) Finally, the acceleration (linear) of the projection M, will be the component of centripetal acceleration of particle P along X axis. f = 2a sin (t + ) ….. (c) From equation (a) and (c), it is evident that f = 2x Proving the fact that the projection M executes SHM along the X axis.

NOTE: the projection of reference particle P upon any diameter will execute SHM. Significance of and Evidently, the projection M repeats its motion accordingly as the reference particle P repeats its motion. The time duration for one complete oscillation for M is same as that for one complete revolution of reference particle P along the circle of reference.

Now, time period = T = 2

Y

'Y A

0M Mx'x

Oa2

t

0P

P

P P

P P

M M


[∵ is the angular speed of P] The phase of the particle (t + ) is the angle made by the radius vector OP with the axis OY’. As the particle P goes on rotating along the circle, the phase of the projection M goes on varying uniformly from 0 to 2, then from 2 to 4 (which can be treated as 0 to 2) and again from 4 to 6 (which can again be treated as 0 to 2)and so on. Even though, the term “phase” is Concerned with particle executing SHM (here projection M) but still it is worthwhile to associate it with the reference particle. Similarly, the term angular frequency () can also be better understood and interpreted in term of the reference particle. Therefore, whenever, we come across problems related to phase difference, it is always preferable and wise to think in terms of the corresponding circle of reference and its reference particle. Things get more easier and more lucid. Illustration 11: A small particle moves along horizontal circle of radius 5 cm with a uniform angular speed of 6

rad/sec. The circle stands in front of a vertical wall, whereon a normal beam of light casts a sharp shadow. What is the maximum speed and maximum acceleration of the shadow? Also find the range between whose limits the shadow moves.

Sol. The situation is similar to the one discussed above. The shadow executes SHM upon the wall. The maximum speed vmax = a = 6 5 = 30 cm/s

And the maximum acceleration max = 2a = (6)2 5 = 1802 cm/s2 The shadow moves along a straight line over the projection of circle upon the wall and its range is the

diameter i.e., 10 cm

IN CHAPTER EXERCISE – 1 1. A particle is executing simple harmonic motion according to the equation

x 5sin t where x is in cm. How long will the particle take to move from the position of equilibrium to the

position of maximum displacement? 2. A body oscillates with simple harmonic motion of amplitude 4 cm and a frequency of 5 Hz. At time

t = 0, the body is at equilibrium position (x =0 ) and moving in positive direction. Write down the equation of simple harmonic motion.

3. A particle executes a SHM of period second and amplitude 2 cm. Find its acceleration when it is

(i) at the maximum displacement from the mean position and (ii) at 1 cm from the mean position. 4. The displacement x (in metre) of an oscillating particle varies with time t (in second) according to

the equation

x 0.02cos 0.5 t3

Calculate (a) amplitude of oscillation (b) time period of oscillation (c) maximum velocity of particle (d) maximum acceleration of particle.


ENERGY ASSOCIATED WITH A PARTICLE EXECUTING SHM Since movement of the particle is involved in SHM, obviously the particle may have kinetic energy with it. Again, since, restoring forces act along the center of oscillation, work is done upon or against the particle, which gets stored in it in the form potential energy, and hence, it can have potential energy too, with it Thus, in general, a particle executing SHM, can be associated with two forms of mechanical energy viz kinetic and potential. (a) Kinetic Energy: If ‘m’ be the mass of the particle executing SHM and v its instantaneous velocity.

Then, K.E.(K) = 2

1mv2

= 2

1m [a cos (t + )]2 [From equation (5)]

= 2

1ma22cos2 (t + ) … (7)

The above equation expresses the kinetic energy in terms of time t. From equation (3), the K.E. of the particle can also be expressed in terms of displacement x.

K = 2

1m [2 ( a 2 x2)] …. (8)

The above equation reveal that the velocity (and hence the kinetic energy) keep on changing with time/displacement and is maximum at the mean position(wa) and minimum (zero) at extreme position. (b) Potential Energy: If ‘’ be the acceleration in terms of displacement x, then we have = 2x Multiplying both sides by mass ‘m’ of the particle executing SHM, we get

Force F = (2m)

x

Since the force acting is a variable one, let us find the work done by the force, between the displacement y to y + dy (where dy is the elemental change in displacement.)

dW = F .

dy m2ydy

Total work done by the force between displacement y = 0 to y = x, can obtained by integrating

W = x

0 m2y dy = m2

2

x 2

We know that, the increase in potential energy from y = 0 to y = x, is the negative of the work done by the forces

U(x) U(0) = W = 2

1m2x2

Where U (x) denotes the P.E. of the system at a displacement x for the particle. Taking u(0) to be zero, we get

U (x) = 2

1m2x2 …. (9)

For equation (4), the displacement ‘x’ can substituted in terms of t

U = 2

1ma22 sin2 (t + ) …. (10)

The above equation reveal that the potential energy of the system, varies with time/displacement with minimum (zero,

assume value) at the mean position and maximum

22ma

2

1at the extreme position.

Total mechanical energy (E) can be obtain by adding K and u both in term of time (t) or both in term of displacement (x).

E = K + U = 2

1ma22 ….. (11)

The above equation reveals that the total energy (E) is independent of time or position of the particle and is always constant.


22

2

1 ma

4

22

ma

222 /ma

O x

K U

E

a 0.707a 0.707antdisplaceme

x

a

Note: In deriving the above expression for P.E. or T.E at mean position is assumed to be zero However if the system has some non zero P.E. at mean position, it should be taken into account to determine P.E. and T.E. The expression for K.E. is independent of the choice of P.E. at mean position. Figure below shows the graphical variation of K.E., P.E. and T.E. with displacement. The curves of K and U intersect at ‘x’ given by

22222 xm2

1xam

2

1

2x2 = a2 x = 0.707 a and the corresponding K or U will be

K = U = 2

1[E] =

4

1m2a2

Illustration 12: Find the position(s) of a particle executing SHM, when its K.E. is 4

1th its T.E.

Sol. From equation (8) K = 2

1m2 (a2 x2)

And from equation (11) E = 2

1ma22

From the problem K = 4

E or 4K = E

22222 am2

1xam

2

14

4a2 4x2 = a2

4x2 = 3a2 x = 2

a3

Illustration 13: Find the ratio of K.E. to P.E. for a particle executing SHM, when it is midway between the mean

position and extreme position.

Sol. Given x = 2

a

Now

2

22

22

222

x

xa

xm21

xam21

U

K

1

3

4

a

4

aa

U

K2

22

Illustration 14: What fraction of total energy is carried by a particle executing SHM, in the from of kinetic energy

when the displacement of the particle is th4

1 the maximum displacement, from one of its extreme

position? Sol. Displacement of the particle from its mean position is

a sayx4

a3

4

a

Now, fraction of energy the particle in the form of K.E.

2

22

222

a

x1

am2

1

xam2

1

U

K


16

7

4

31

E

K2

Illustration 15: A particle of mass 200 g is executing SHM with an amplitude 20 cm. If its kinetic energy be

32 103 joule initially, determine its displacement equation, assuming the phase constant to be 4

Sol. Let be the angular frequency, for the SHM, then displacement equation can be written as x = 20 sin (t + /4) where t is in sec. and x is in cm.

Now, initially x = 20 sin

4 [Putting t = 0]

2

20= x x2 =

2

400 = 200

Now, kinetic energy at a displacement x, will be

K = 2

1m2 (a2 x2)

32 103 2

1

1000

200 2 (400 200) 102

2 = 210200200

232

=

2

8 = 4 rad/sec.

Thus, the displacement equation will be x = 20 sin

4

t4

Illustration 16: A particle SHM with force constant 2 106 N/m and amplitude 1 cm, and has a total mechanical energy of 200 J. Find the following (a) maximum K.E., (b) Maximum P.E., (c) Minimum K.E.,

Sol. given k = m2 = 2 106 ; a = 1 102 m

(a) Maximum K.E. = 2

1m2a2 [ |vmax| = a]

= 2

1 2 106 1 104 = 100 J

(b) From E = K + U E = Umax [when K = 0 ; at the extreme position] Maximum potential energy Umax = 200 J

(c) Minimum kinetic energy = 2

1m (vmin)2

= 0 [at the extreme portion] (d) Minimum Potential energy Umin = E Kmax = 200 100 = 100 J

NOTE: The particle’s potential energy is not zero at mean position.

IN CHAPTER EXERCISE – 2

1. A particle of mass 0.2 kg executes a SHM of amplitude 2 cm and period 6 second. Find (i) the total mechanical energy at any instant (ii) kinetic and potential energies when displacement by 1 cm.

2. A particle executes SHM of amplitude A. (i) At what distance from the mean position is its kinetic energy equal to potential energy? (ii) At what points is its speed half the maximum speed?

3. A 2 kg body panel of a car oscillates with a frequency of 2 Hz and amplitude 2.5 cm. If the oscillations are assumed to be simple harmonic and undamped, calculate


(a) the maximum velocity of the panel

(b) the total energy of the panel

(c) the maximum potential energy of the panel, and (d) the kinetic energy of the panel 1.0 cm from its equilibrium position.

4. A body executing linear SHM has a velocity of 0.03 ms-1 when its displacement is 0.04 m and a velocity of 0.04 ms-1 when its displacement is 0.03 m. (a) Find the amplitude and period of the oscillation (b) If the mass of the body is 50 × 10-3 kg, calculate the total energy of oscillation.

SIMPLE PENDULUM A small heavy spherical bob connected, to one end of a light inextensible and flexible string, with its other end fixed to a rigid and unyielding support, constitutes a simple pendulum. Let P be the position of the bob at any instant, when the string makes an angle , with the vertical. (Figure) If ‘l’ be the length of the string then, the net unbalanced force on the bob is F = mg sin . [where m is the mass of the bob]. Assuming to be very small sin Net unbalanced force F = mg ()

Now, if ‘x’ be the displacement of the bob from the mean position C, then = l

x

l

xmgF

Acceleration f = l

)x(g

m

F

f ( x), which proves that the bob executes SHM for small amplitudes (less than 4º) Comparing with f = 2 x,

we get 2 = l

g =

l

g

Time period of oscillation T = g

l

22

Alternatively : The torque acting about the point of suspension is I = ml2 From T = I, we get

mgl sin = ml2 = l

sing

l

)(g

dt

d2

2

Comparing with 2

2

dt

d = 2 ()

= l

g or T = 2

g

l

NOTE: Here ‘l’ is the “effective length” of the simple pendulum, measured from the point of suspension to the centre of gravity of the bob.

A Few Important Facts Regarding Simple Pendulum (a) Change of time period T due to a change in effective length. [Law of lengths] T l [other things remaining constant] (i) If a child sitting on a swing stands up, then the effective length l decreases, due to an uplift of the centre of gravity of the system. This reduces the time period consequently. (ii) If a hollow sphere filled with a liquid, and having a hole at its base, be made the bob of a simple pendulum, the time period, initially increases, due to an increase in the effective length as the centre of gravity gets lowered due to


the outflow of liquid from the sphere. The time period once again, gets restored, when the whole liquid flows out, due to the fact that the centre of gravity once again, comes to the centre of the sphere. (iii) if the string be elastic with Young’s modulus Y, then the length of pendulum will be l + l [where l is the original length and l, increase in length]

From Y = l

l

ΔA

MgY

strain

stress

[where M is the mass of the bob]

l

l =

AY

Mg

Net length = l

AY

Mg11 l

l

l

Now, Time period T = 2

AY

Mg1

g

l

Now, since Mg < < AY

AY

Mg < < 1

T = 2 g

l

AY2

Mg1 [From Binomial expansion]

Obviously, the time period increases, in this case. (iv) If the effective length be comparable to the radius of earth. (Figure) The net unbalanced force on the bob = mg sin ( + ) F = mg( + ) [Assuming ( + ) to be small]

R

xxmgF

l

= mg xR

11

l

or acc. f = g xR

11

l

which is of the form of x

2f , indicating that the bob, still executes SHM and its time period is not infinity. Comparing the above equation with std. form

2 = g

R

11

l

Time period T = 2

R

11g

1

l

Putting l = R, we can determine, the corresponding time period as

T = 2g2

R 1 hr

If l be too large i.e., l 1/l 0

T = 2g

R 84.6 min.

If l be too small compared to earth’s radius, then

l

1

R

1

Neglecting R

1 in comparison to

l

1

T = 2g

l as derived earlier.


(b) change of time period with a change in acceleration due to gravity. (Law of gravity)

T g

1 [other things remaining constant]

(i) If a simple pendulum be taken to higher altitudes, its time period increases due to a decrease in the value of g. Consider a place P at height ‘h’ above the earth’s surface. (Figure) The value of g at that place g’ (say) is given by :

g' = 2

2

)hR(

gR

where g = acc. due to gravity on the earth’s surface.

Time period T = 2'g

l

22

2

gR

)hR( l

T = 2

R

h1

g2

gR

hR ll

Now, if h < < R or R

h < < 1

Then T = 2 2

Rh

1g

l = 2

R

h21g

l

(ii) If a simple pendulum be taken to a place below the earth’s surface, the time period increases as a consequence of decrease in the value of acceleration due to gravity. Consider a place P, at a depth ‘d’ below the earth’s surface, (figure) the acc. due to gravity at P is given by :

g’ = g

R

dR [where g is the acc. due to gravity on the earth’s surface]

Now, time period T = 2 'g

l 2

d)g(R

R

l

(iii) If a simple pendulum at equator, be shifted to poles, the time period decreases on account of an increase in the value of ‘g’. (acc. due to gravity is more at poles compared to equator). (iv) The time period of a simple pendulum in a satellite within a freely falling lift is infinite (or the pendulum does not oscillate due to the absence of any restoring force, tending the bob to bring back to the mean position), on account of the fact that the value of g is zero. (v) For a simple pendulum oscillating in a lift moving up with an acceleration a or down with a retardation a the effective value of g (say g’) will be g’ = g + a

T = 2 )ag(

l ;

The time period always decreases. For a simple pendulum, oscillating in a lift moving down with an acceleration a or moving up with a retardation a, the effective value of g (say g') will be g' = (g a) down if g > a or (a g) up if a > g. (Figure)

EARTH

P

R

O

h

P

d-R

O

d

)ag(T

l2

)ga(T

l2

a a


For a simple pendulum oscillating in a vehicle accelerating with an acceleration ‘a’, the time period decreases due to an increase in the value of ‘g’. Let us first determine the equilibrium position of the string (see figure). If be the angle made by string with vertical, then : T cos = mg and T sin = ma

Dividing tan = g

a

= tan1

g

a

(opposite to the direction of acceleration of vehicle)

Effective external force = m 22 ga .

Effective acceleration g' = (a2 + g2)1/2

Time period T = 2 'g

l = 2

2/122 )ag(

l

(iv) If the bob of a simple pendulum is acted upon by additional forces of electrical, magnetic, buoyant, gravitational origin, the time period will increase or decrease according as the net acceleration in the downward direction is decreased or increased.

Illustration 17: Show that the maximum tension in the string of a simple pendulum with angular amplitude 0 is given by mg )1( 2

0 . Hence find the maximum mass of bob, that can be used, if the breaking load

of the string be F. Sol. Evidently, the tension is maximum, at the lowest position of the bob. (figure)

Tmax = mg + l

2maxmv

........ (1)

where vmax is the maximum velocity of the bob (attained in the mean position)

|vmax| = a

2maxv = a2 2 (0l)2 2

0

2g

l(gl)

[where a = linear amplitude and = ang. Freq.]

Substituting for 2maxv in eqn. (1)

Tmax = mg + )1(mg)g(m 2

0

20

l

l

If M be the maximum mass, that can be employed for the bob.

Then, F = Mg )1( 20

M = )1(g

F20

gatan 1

A

B

Cvertical

a

0 0

Tv

mg l

2mv


qE

mg

qE

mg

)a( )b(

( mea

n po

siti

on)

E T

qE

mg

Illustration 18: A simple pendulum of length l, is made to oscillate with a solid sphere (density D), in a non viscous fluid of density ; find the period of oscillation. [D > ]

Sol. Net force on the bob will be (D )Vg down). Net Acceleration downward (Figure)

f = gD

1DV

Vg)D(

Time period T = 2

D1g

l

The time period increases, due to a decrease in effective value of g. Illustration 19 :A simple pendulum consists of a small sphere of mass m, suspended by a thread of length L. The

sphere carries a negative charge ‘q’. The pendulum is placed in a uniform electric field of strength E directed (i) vertically upwards, (ii) vertically downwards, and (iii) horizontal. Find the time period in each case.

Sol. (i) The net force downward F = mg + qE Net acceleration downward

m

qEg

m

Ff

Time period T =

m

qEg

L2

NOTE: The electric force experienced by a body carrying a negative charge is opposite to the direction of electric field.

(ii) The net force downward = mg qE [Assuming mg > qE] [Figure (a)]

Acceleration f = g

m

qE

Time period T =

m

qEg

L2

The net force upward = qE mg [Assuming mg < qE] [Fig. (b)]

Acceleration f =

m

qE g

Time period T =

gm

qE

L2

(iii) If '' be the angle made by the string with the vertical (Figure) Then T cos = mg T sin = qE

mg

qEtan

mg

qEtan 1

Net force on the bob = 22 )qE()mg(

Net acceleration f = 2

2

m

qEg


Time period T = 2/12

2

m

qEg

L2

Illustration 20: A simple pendulum, of length l, installed between two inclined walls, each making an angle

with the vertical, oscillates with a maximum speed of u, colliding elastically with the wall. Find the time period of oscillation. For the above problem, how is related to u ?

Sol. Had there been no walls, the linear amplitude of oscillation would be a = u

[From |vmax| = a]

But since = l

g

a = ug

l

Now, let the equation of motion for the pendulum be = 0 sin t [where 0 = angular amplitude]

The time taken by the pendulum to move from O (mean position) to A (extreme position), can be obtained as

=

t

gsin

g

u

ll

sinu

l

l

gt

g t =

u

ll gsin

g1

u

ll gsin

gt 1

Evidently, the period of oscillation T = 4t

T = 4

u

ll gsin

g1

Now, since, bob collides with walls, the angular amplitude should exceed the value

i.e., lg

u or u > lg

This is the required relation

NOTE: If the bob just misses the collision with wall, Then, in the limiting case, angular amplitude =

Then T = 4g

l sin1(1)

T = 4g

l

2 = 2

g

l

These prove the validity of above derivation. Second’s Pendulum : It is pendulum with period of oscillation 2 sec or period of swing 1 sec. Length of a

second’s pendulum at mean sea level, may be calculated as follows.

From T = 2g

l we have l =

2

2

4

gT

Putting T = 2 and taking 2 g l 1 m


NOTE: A second’s pendulum always bears a time period of 2 sec. however, the length changes with a change in acceleration due to gravity.

Loss or Gain of Time by Pendulum Clocks Pendulum Clocks gain time, if the time period of the pendulum decreases (due to decrease in length in winter or due to increase in acc. due to gravity, as when taken from equator to poles). For example, let the time period of a second’s pendulum (originally) decrease to 1.99 sec. Now, each time, it oscillates, it counts 2 rather than 1.99 s. Thus, after a considerable no. of oscillations, the time counted or shown by the clock is more than the actual time, and consequently, is goes fast. Similarly, pendulum clocks loose time and go slow, when their time period increases. Gain or loss of time due to a change in surrounding temperature : Let T be the time period of clock pendulum of effective length l, showing the correct time.

Then T = 2g

l …….. (1)

Now, let be the change in temperature of the surrounding, the new length will be l' = l (1 + ) [where is the coefficient of linear expansion of the material of the pendulum] The changed time period

T' = 2g

l'

= 2g

αΔθ)(1l …….. (2)

Dividing equation (2) by (1)

2/1)1()1T

'T

2

11 [Expanding by binomial expansion]

T' = T

2

11

Also,

2

1

T

T'T

i.e., Relative change in time period = 2

1

or, time lost or gained each second is = 2

1

NOTE: Unless otherwise stated, adopt the time period of clock pendulum as 2 sec.

Illustration 21: A clock with an iron pendulum keeps correct time at 20ºC, how much time will it loose or gain, in a

day, if the atmospheric temperature changes to 40ºC. for iron = 0.000012/ºC) Sol. The relative change in time (i.e., time lost or gained per swing i.e., in 1 sec is given by

2

1

2

1

T

T

12 106 (40 20)

12 105 s Total time lost per day = 12 105 86400 = 12 864 103 = 10.37 sec.

Illustration 22: The pendulum of a certain clock has a time period 2.02 sec. How fast or slow does the clock run in an interval of one week?

Sol. Since the correct time period for a clock pendulum is 2 sec. Therefore, for every 2.02 sec. (one oscillation period) of actual time elapsed, it would count only 2 sec.

In other words, as successive oscillations are repeated it goes on loosing time, and would run slow.


Loss of time per oscillation ( 2 sec) = 2.02 2 = 0.02 sec. Loss of time per sec = 0.01 sec. Time lost in 1 week = 0.01 86400 7 = 6048 sec = 108 min. Illustration 23: The linear displacement ‘x’ of a simple pendulum’s bob from its mean position varies as x = a sin

(0.707 t) where a is a constant and t is in sec. Find its length (take g = 2 m/s2). Sol. Since the simple pendulum executes SHM. Therefore comparing the displacement equation with standard

form x = a sin (t), we get

= 2

rad/sec.

Also from T = 2g

l =

2

we have 2 = l

g

l

g

2

2

l = 2m

Illustration 24: A simple pendulum of length l metre, suspended from the ceiling of a lift executes oscillation with a

period of 6.25 sec. If the lift is moving up with an acceleration ‘a’ find a (take g = 2 m/s2). Sol. If ‘a’ be the acceleration of the lift upwards then, net acceleration on the bob downward, relative to the

lift is g + a

T = 2ag

l 6.25 = 42

a

12

2 + a = 2

25

16 a = 2

25

9

Thus, the lift is retarding with a retardation of 2

25

9 m/s2.


1. The acceleration due to gravity on the surface of moon is 1.6 ms-2. What is the time period of a simple pendulum on moon if time period on earth is 3.4 second? Given: acceleration due to gravity on earth = 10 ms-2.

2. The bob of a vibrating simple pendulum is made of ice. How will the time period change when the

ice starts melting? 3. The bob of a simple pendulum is positively charged. The pendulum is made to oscillate over a

negatively charged plate. How shall the time period change as compared to the natural time period of the simple pendulum?

4. In a pendulum oscillation, the amplitude is 0.05 m and the period 2.0 second. Calculate the

maximum velocity. 5. A simple pendulum makes 50 oscillations in 100 second. The amplitude of an oscillation is 2 cm.

Calculate the particle velocity amplitude and acceleration amplitude.


SPRING MASS SYSTEM (a) Horizontal oscillation of a mass attached to an ideal spring Consider a block of mass m, attached to one end of a massless spring,

obeying Hooke’s law, with its other end attached to a rigid support. The block rests on a horizontal frictionless surface. (Figure)

Let the block be slightly pulled towards right and left free to oscillate. It executes SHM. Consider the position of the block, at any instant,

when its displacement is ‘x’ from the mean position.

The net force acting upon the block is the restoring force due to spring, (towards left).

Thus F = xk [where ‘k’ is the force constant]

or f

= )x(m

k

m

F

Comparing with the standard from F = x2

We have m

k2

Now, time period of motion

T = k

m2

2

Energy Conservation in Spring Oscillation Let ‘x0’ be the maximum compression ( or extension) in the spring. In such situation

K = 20

2 kx2

1U'0)0(m

2

1

and E = K + 20kx

2

1U …..(a)

Now, if v0 be the maximum speed, of the block, when at mean position, then

K = 0)0(K2

1U;mv

2

1 20

E = K + U = 20mv

2

1 …. (b)

But, as the block moves from its mean position, to either of its extreme position, it utilizes, its kinetic energy

2

0mv2

1 to elongate or compress; the spring, which gets stored in the spring in the from of potential

energy

2

0xk2

1.

Thus, from (a) and (b), it is evident at the total energy of the system (i.e., spring + block) remains constant. NOTE: Oscillation of a spring pendulum is independent of the value of ‘g’ at the place Horizontal oscillations, of a block attached to a spring, on a rough surface. Consider a block of mass ‘m’, attached to a spring of force constant k, oscillation on a rough horizontal surface with coefficient of friction ‘’. Let a0 be the initial compression, of the spring and a1 be the maximum extension, for the first time. From energy consideration

1021

20 aa)mg(ka

2

1ak

2

1

101010 aamgaaaa2

k a0 a1 =

k

mg2

C P

KxF x

0a 1a


(b) Vertical oscillation of a block attached to an ideal spring. Consider a block of mass m attached to one end of an ideal spring, with its other end fixed to a rigid support. Let ‘y0’ be the initial extension in the spring. (Figure)

If ‘k’ be the force constant of the spring, then for equilibrium (vertical) of the block. ky0 = mg (numerically) Now, if the block be slightly displaced from its mean position it starts executing SHM. Let ‘y’ be the displacement of the block, at any instant from its mean position

(downwards). The net force acting on the block upward will be F = k (y0 + y) mg

or ykF

or acceleration F k

f ym m

Comparing the above equation with the standard from F = 2 y , it is evident that the block executes SHM,

with an angular frequency = m

k

The period of oscillation T =

22

k

m

NOTE: Even the vertical spring oscillation is also independent of the value of ‘g’ at that place. (c) Horizontal spring oscillation, taking into consideration, the mass of the spring. As the block oscillates, so does the spring. However the velocity at each point of the spring is not the same. In

the elongated (or compressed) state, the displacement of each section can be assumed to be linearly proportional from their respective distance from the fixed end.

Thus, if ‘y’ be the displacement of the block (i.e., the end of the spring connected to the block) at any instant,

then the displacement of a section distance ‘x’ from the fixed end will be L

yx where L is the length of the

spring; similarity if v be the velocity of block then v = dt

dy (Figure)

Accordingly the velocity v’ of the aforementioned section will be

L

xv

dt

dy

L

xv '

Now, kinetic energy of the entire spring can be obtained as the sum (integral) of the kinetic energies of each section.

Thus, total K.E. of the spring =

2,vL

dx'M

2

1

[Here M ' is the mass of the spring and dx, the length of the elemental section.]

= L

'M

2

1

L

0

3L

03

2

2

22

3

x

L

v'M

2

1

L

dxxv

=

3

v'M

2

1 2

Total kinetic energy of the spring + block system = 22

Mv2

1

3

vM'

2

1

[where M is the mass of the

block] Now, the total mechanical energy of the system is given by E = K + U

= 2

1

3

v'M 2

+ 2

1Mv2 +

2

1ky2

Since the total mechanical energy should remain conserved

dt

dE = 0

0y

y

C

(mea

n po

siti

on)

)positionmean(

v

x

y

dx


0dt

dyy2

2

k

dt

vdv2

2

M

dt

vdv2

6

'M

v

dt

dy0ky

dt

dvM

3

'M

Again, since 2

2

dt

yd

dt

dy

dt

d

dt

dv

M

3

'My

dt

yd2

= 0

Comparing with the standard form 2

2

dt

xd + 2y = 0

We find, that the complete system executes SHM,

with an angular frequency =

M

3

'M

k

Time period of oscillations T = 2k

M3

'M

NOTE: (1) The above equation can be rewritten as T =2k

M eff where Meff =

M

3

'M

(2) The oscillations of a block attached to a spring, remains SHM only for small elongation (or

compression) when the restoring forces generated in the spring remains linear. For large deformations in spring, the restoring forces become nonlinear.

(3) Unless, otherwise specified, a spring should always be treated as ideal and massless.

Composite spring oscillations.

(a) When springs are connected in series: Consider three springs with stiffness constants k1, k2 and k3 respectively,

If x1, x2 and x3 be the respective elongations of the three springs, then, the displacement of the block x is given by

x = x1 + x2 + x3 ……. (a)

Now, if F be the restoring force acting along each of the spring

Then F = k1x1 = k2x2 = k3x3 …….. (b)

Now, the net force acting on the block will also be F

From (b), x1 = 1k

F ; x2 =

2k

F and x3 =

3k

F

Substituting them in equation (a)

x = 321 k

1

k

1

k

1

or

321 k

F

k

F

k

FF

effK

F

M

1k

x

3x2x1x

2k 3k


If m be the mass of the block, then acceleration

321 k

1

k

1

k

1m

x

m

F

f

Replacing 321 k

F

k

F

k

F with

effK

1 we have

m

)x(keff

f

which proves the fact that the oscillations are SHM, with angular frequency = m

k eff

Time period of oscillation = T = effk

m2

2

Where eff 1 2 3

1 1 1 1

k k k k

(b) When springs are connected in parallel:

It can be shown that the oscillations of the block are SHM with a time period T given by T = 2effk

m

where keff = k1 + k2 + k3 (c) Figure below shows a special arrangement of two springs virtually appearing to be in series ; however they are not. If x

be the displacement of the block at any instant then, the net force acting on it

xkxkFFF 2121

= (k1 + k2) x

The effective spring constant for the above set up is keff = k1 + k2

Time period T = 2effk

m

Illustration 25 : A small block of mass m is attached to three springs with respective stiffness constants k1, k2 and k3 (see fig.). Find the period of oscillations, if the block is slightly pressed against the spring with stiffness constant k3. Will your answer be the same, if the block be pressed against the spring with stiffness constant k1 ? Sol. Figure below shows, the respective elongations occurring in the springs of stiffness constant k1 and k2, when

the spring of stiffness constant k3 is compressed by ‘x’. Net restoring forces acting on the block along the spring of stiffness constant k3 will be

M

1k

2k

3k

M1k 2k

xkF 11

xkF 22 x

º90

º45

1k

2k

3k


1FF

cos 45º + 2F

cos 45º + F3

xk

2

1xk

2

1xk 321

F

= xk2

k

2

k3

21

Thus, the effective stiffness constant is given by

keff = 321 k

2

kk

Time period of oscillations T = 2effk

m

If the block compressed against spring of spring constant k1, the net force will be (see Figure)

F

= 23

231 )º45sinF()º45cosFF(

= 2

32

31 2

xk

2

xkxk

F

= )x(2

k

2

kk

2/1

32

31

Effective stiffness constant

Keff =

2/123

23

1 2

k

2

kk

Illustration 26 : Figure shows three springs A, B and C with force constant k1, k2 and k3

respectively, connected to a block of mass m, passing through a light pulley. Assuming all other things to be massless, find the period of oscillations.

Sol. Evidently, springs A and B are in series, with effective spring constant k' = 21

21

kk

kk

which in turn is in parallel with the spring C.

Effective spring constant keff = 321

21 kkk

kk

Time period T = 2effk

m

2313221

21

kkkkkk

)kk(m

Alternative: (from energy considerations).

Let y0 be the initial compression in the spring C and, for the sake of simplicity, let the spring A and B be in the relaxed position, when the block is at rest (Figure)

For vertical equilibrium of the block mg = k3y0.

Let ‘y’ be the displacement of the block (below its mean position), which would also be the additional compression in the spring C.

Let y1 and y2 be the respective elongations of the spring, A and B respectively.

Since total extension in spring A

= net compression in springs B and C

y1 = y y2 y = y1 + y2 …….. (1)

º45cosxkF 22

º45cosx

º45cosx

xkF 33

F 1=

k 1 (

x co

s

xkF 11

º45cosxkF 33

º45cosx

A B

C

1k

2k

3k

m


Also, since the tension in the two ends of the string (passing through the pulley) should be same therefore, k1y1 = k2y2 …….. (2)

From (1) and (2) y1 =

21

2

kk

yk and y2 =

21

1

kk

yk

Total mechanical energy will be

mgymv2

1)yy(k

2

1yk

2

1Eyk

2

1E 22

03222

211

Since, total mechanical energy remains constant, during oscillation

0dt

dE

i.e., 0mgy2

mv)yy(

2

k

kk

yk

2

k

kk

yk

2

k

dt

d 22

03

2

21

122

21

21

0dt

mgdy

dt

mvdv

dt

dy)yy(K

dt

dy

)kk(

ykk

dt

dy

)kk(

ykk032

21

212

221

221

0dt

mvdvyk

)kk(

ykk

)kk(

ykk32

21

212

221

221

032

2

321

21 ykmgandvdt

dy0

dt

ymdyk

kk

kk

or 0ykkk

kk

m

1

dt

yd3

21

212

2

Comparing with the standard form 2

2

dt

yd + 2y = 0 we find that 2 =

21

313221

kk

kkkkkk

m

1

Time period T = 313221

21

kkkkkk

)kk(m2

2

Illustration 27: A horizontal spring block system (of force constant k) and mass M executes SHM with amplitude

A. When the block is passing through its equilibrium position an object of mass m is gently placed on it and the two move together. Find the new amplitude and the minimum coefficient of friction between the two blocks.

Sol. Let the initial angular frequency be and the velocity of the block at its mean position be v.

Then = M

k

Also v = max. vel. of the block

= AM

k …….. (1)

Now, if v' be the combined speed, then, from law of conservation of linear momentum (M + m)v' = Mv

v' = mM

Mv

and, the new angular frequency will be ' = mM

k

Again from |vmax'| = A''

mM

k'A

mM

Mv

[where A' is the new amplitude]

mM

k'A

M

kA

)mM(

M

[From eqn. (1)] A' = A

mM

M

1y

2y

y

(mea

n po

siti

on)

M

m

v


Illustration 28: Two bodies A and B of equal mass are suspended from two separate massless springs of force constant k1 and k2 respectively. If the bodies oscillate vertically such that their maximum velocities are equal, then find the ratio of the amplitudes of A and B.

Sol. If 1 and 2 be the respective angular frequencies of the oscillations related to A and B.

Then 1 = m

k1 and 2 = m

k 2 [where m is the mass of each body A and B]

Now, if a1 and a2 be their respective amplitudes then : |vmax| = a11 = a22

1 2 2

2 1 1

a k

a k

Illustration 29: If a body attached to a spring executes vertical oscillations with time period T, find the new time

period, if the same body be attached to a new arrangement in which the spring is subdivided into four parts of equal length each and all of them connected in parallel.

Sol. Let k be the stiffness constant of the spring and m the mass of the body. Then time period ‘T’ of the oscillations will be given by

T = 2k

m …….. (1)

Now, since the stiffness constant of a spring varies inversely as its length Stiffness constant of each part = 4k Further, the effective stiffness constant, when all of them are connected in parallel will be 4 (4K) = 16k. Now, the new time period

T' = 2k

m

4

2

k16

m

T' = 4

T [From (1)]


1. A body of mass 9 kg is oscillating on a spring of force constant 100 N m-1. Deduce angular frequency.

2. Two springs are joined in series and connected to a mass m as shown n figure. If spring constants are k1 and k2. Calculate the period of oscillation of mass m.

m k2 k1

3. A 0.3 kg mass is connected to a system of two identical springs as shown in figure. Determine th

period of motion for the system. Ignore frictional forces.

k=20

Nm

-1

k=20

Nm

-1

0.3 kg 4. A 2.5 kg collar attached to a spring of force constant 1000 N m-1 slides without friction on a

horizontal rod. The collar is displaced from its equilibrium position by 5.0 cm and released. Calculate (a) the period of oscillation (b) the maximum magnitude of acceleration, and (c) the maximum speed of the collar.


5. A tray of mass 12 kg is supported by two identical spring as shown in figure. When the tray is pressed down slightly and released, it executes SHM with a time period of 1.5 s. What is the force constant of each spring? When a block of mass M is placed on the tray, the period of SHM changes to 3.0 s. What is the mass of the block?

m12kg

ANGULAR SIMPLE HARMONIC MOTION

In this type of SHM, the particle (or body) involved moves to and fro along an arc of a circle. Besides, the motion should be such that:

(a) It is oscillatory. (b) The resultant torque acting on the particle about some fixed axis normal to the plane of motion, is

proportional to its angular displacement from the mean position, and is always direction such that it has a tendency to bring the particle to its mean position.

Thus, if be the resultant torque, and ‘’ be the angular displacement, in any position, then ( ) = k …….. (12) where k is constant of proportionality. Now, if ‘’ be the angular acceleration and I the moment of inertia of the particle about the centre of rotation, Then, = I

=

I

k …….. (13)

( ) Equation (13) corresponds to equation (1) in linear SHM

= I

k or T =

2

= 2k

I

Substituting = 22

2

I

kand

dt

d

in eqn. (13).

Equation for angular SHM can be written as

2

2

dt

d + 2 = 0

Solving the above equation, equations for angular displacement and angular velocity can be obtained as = 0 sin (t + ) …….. (14) and = 0 cos (t + ) …….. (15)

and = 220 …….. (16)

where ‘0’ is the angular amplitude. Notice that the symbol has been used to denote angular velocity of the particle to distinguish it from ‘’

angular frequency. NOTE: Equations corresponding to angular SHM can be obtained from equation pertaining to linear SHM, by

replacing mass by moment of inertia, force by torque, linear displacement by angular displacement etc. Mechanical Energy Associated with Angular SHM (1) Kinetic energy (rotational) (k) If I and be the moment of inertia and angular velocity of the oscillating body,


Then kinetic energy k = 2

1 I 2

= 2

1 20I 2 cos2 (t + ) [In terms of time]

= 2

1I2 ( 2

0 )2 …….. (17) [In terms of angular displacement]

(2) Potential energy (U) Work done by the internal torque, while the particle moves from “mean position” to an angular position ‘’

will be

W =

0 d

W =

0 k d

=

0I2 d

= 2

I 22

Now, change in potential energy = Negative of the work done. Thus,

U () U(0) =

2

I 22

= 2

I 22

If we assume that = 0 (i.e., mean position) corresponds to zero potential energy

Then, U = 2

1I22 =

2

1 20

2I sin2 (t + ) …….. (18)

(3) Total energy (E) E = k + U

= 2

1 I2 ( 2

0 2) + 2

1I22

= 2

1 I22 …….. (19)

Thus, the above equation reveals the fact that the total energy in angular SHM remain constant.

PHYSICAL PENDULUM OR COMPOUND PENDULUM

Any rigid body, suspended about a fixed point and capable of oscillating about a horizontal axis passing through the point, constitutes a physical or compound pendulum. Figure alongside shows a rigid body (of any shape) suspended about a fixed point O (the centre of suspension) capable of rotating about a horizontal axis through O. Let G be the centre of gravity at a distance l from the point of suspension O. If ‘R’ be the vertical reaction upwards at the support O, then for vertical equilibrium, we have R = mg When at rest, the line of action of R and mg is the same and hence, there is no net torque on the body. If the body be given a slight angular displacement about O, it starts oscillating, the oscillations being angular SHM. Consider the position of the body, at any instant (during oscillatory motion) when the line joining O and G makes an angle ‘’ with the vertical. The net torque on the body about the point O will be given by = mgl sin IF I be moment of inertia of the body about O, then

O

G

R

mgvertical

l


= I 2

2

dt

d [where

2

2

dt

d = angular acceleration]

I2

2

dt

d = mg l sin

[The ve sign, here indicates that the direction (or sense) of and are opposite] [ for small angle sin = ]

2

2

dt

d + 0

I

mg

l

Which reveals the fact that, the body performs angular SHM with angular frequency

= I

mgl

Now, if T be the time period for each oscillation, then

T = 2

= 2lmg

I …….. (20)

If K be the radius of gyration of the body about a parallel axis through the centre of gravity then, I = mK2 + ml2

T = 2l

l

g

K 22 …….. (21)

If L be the length of an equivalent simple pendulum (i.e., a pendulum whose time period is same as that of the compound pendulum)

then T = 2g

L

Comparing the above equation with equation (21) we get L = ll

2K

Thus, the length of equivalent simple pendulum is l

2K more than the distance between the point of suspension O and

the centre of gravity G.

If we choose a point O', on the line OG produced such that GO' = l

2K, then O' is known as the centre of oscillation.

Suppose we shift the point of suspension from O to O'.

Then time period T' = 2'g

'K 22

l

l [From eqn. 21]

where l’ is the distance of O' (new point of suspension) from centre of gravity G.

But GO' = l

2K l' =

l

2K

T' = 2

l

l

l

l2

222

2

222

Kg

KK

2K

g

KK

= 2

22

g

)K(2

l

ll

= l

l

g

K2

22 which is same as equation (19)

T' = T

O

'O

G

l

2K

l


Thus, it is evident that, the time period remains the same for both centre of suspension and centre of oscillation i.e., both are interchangeable. Illustration 30: A uniform rod of length l and mass m, is suspended by one end and is displaced slightly, to undergo

angular SHM. Find its time period. Sol. Consider the rod, in a position making an angle ‘’ with the vertical at any instant. (figure) torque acting on the rod about the point of suspension

= mg 2

1 sin

= l

lmg [for small angle sin ]

Now, angular acceleration 22

2

m

3

Idt

d

l

= 2

mg

dt

d

3

m2

22

ll

or 02

g3

dt

d2

2

l

Comparing with the standard form 0dt

d 22

2

We get = l2

g3 Time period

g3

22

2T

l

Illustration 31: A square lamina of edge ‘a’ is suspended through a hole punched on a diagonal at a distance of

22

a from its centre. Find the period of oscillations.

Sol. Consider, the oscillating lamina (of mass m say) at any position making an angle ‘’ with the vertical. Net torque acting on the lamina about the point of suspension O, will be (Figure)

= mg 22

a sin = mg

22

a ()

Now, the moment of inertia of the lamina about an axis through O, and normal to its plane will be

I = 22

22

am

6

ma

= 2ma24

7

Since = 2

2

dt

Id

22

mga

dt

d

24

ma72

22

22

2

dt

d

we get 2 = 27

g12

Time period T = g3

27

g12

272

2

Minimum Period of Oscillation.

From equation (21), the time period T for a compound pendulum is given by T = g

K

2

2

ll

which can be

rewritten as

2l

mgvertical

mgR

O

.G.C

mg

vertical


T =

K2K2)(

K

g

12 2

2

ll

=

K2)(

K

g

12

2

ll

2K is constant, therefore T will be min. when 2

)(K

l

l is also min. Since the min. value of a square can be

zero. K = l Thus, the time period of oscillations is minimum when the distance of the centre of suspension from the centre of gravity is same as the radius of gyration of the body about a parallel axis through C.G.

Also, Tmin = 2g

k2 …….. (22)

Illustration 32: A uniform circular disc of radius R, is suspended about a point (not its centre such that it can oscillate about a horizontal axis through that point. What is the distance of this point from the centre, for the time period to be minimum? What is the minimum time period?

Sol. The radius of gyration for a disc about its centre of gravity (i.e., geometrical centre) is given by

k = 2

R

M2

MR

M

Ik

2

This is the required distance

Also, Tmin = 22g

R2 [See equation 22]

= 2g

2R

Illustration 33 : Find the frequency of oscillations of a uniform thin ring of radius ‘R’ suspended about a point

on its circumference.

Sol. From equation (20) T = 2lmg

I

Here I = 2mR2, l = R

frequency n = R2

g

2

1

mR2

mgR

2

12

TORSIONAL PENDULUM [OPTIONAL]

A rigid body suspended by means of a wire, with its one end fixed rigidly to the ceiling, and about which the body can perform horizontal angular oscillations, is known as a torsional pendulum. Figure shows a body attached to a wire OO', whose one end O, is fixed to the ceiling. When the body is at rest, there is no torque on the body. If the body be given a slight “twist”, and left free, it starts performing angular SHM. Let O'P be horizontal line chosen on the body at rest. Further let O'P' be its position at any instant, during oscillations. Since the upper end O of the wire is fixed, and the lower end O', is rotated by an angle . We say that the wire has a twist, ‘’. Due to this twist,

R

.G.C

O

'O

P 'P


a torque (restoring) is developed in the wire due to elastic forces, which tend to bring the wire and the body back to a position = 0 (i.e., the mean position). This restoring torque is directly proportional to the twist, but opposite in sense. Then ( ) or = k where k is a positive constant of proportionality known as torsional constant (for the wire). If I be the moment of inertia of the body about the wire, then

= 2

2

dt

dI

2

2

dt

dI = k 0

I

k

dt

d2

2

which proves the fact that, the body executes angular SHM, with an angular frequency.

= I

k

Time period T = k

I2

2

…….. (23)

Illustration 34 : A circular disc of mass m = 500 gm and radius r = 5 cm is suspended by means of a wire of

torsional constant k = 0.36 N – m/radian. Find the time period of oscillations. Sol. Moment of inertia of the disc about the wire

= 2

mr 2

From equation (23), time period of the torsional pendulum

T = k

I2

k2

mr2

2

Substituting the values of m, r and k

we get T = 22

22

10362

)105(

2

1

= 2

3620

52

364

1025 2

T = 72

sec.

Illustration 35 : A torsional pendulum consists of a square lamina, suspended by means of a wire with torsional

constant k = 0.25 N - m/rad. If the time period of oscillations be 2.5 sec, find the moment of inertia I of the square lamina, about the wire.

Sol. The time period for oscillations of a torsional pendulum is given by T = k

I2

or I = 2

2

2

2

)14.3(4

25.0)5.2(

4

kT

= 0.04 kg – m2

GENERAL METHOD TO SHOW THAT A GIVEN BODY (OR A PARTICLE) EXECUTES SHM AND TO FIND ITS TIME PERIOD.

1. FORCE/ TORQUE METHOD

Step 1: Establish the equilibrium condition (if any) for the body at its mean position. netF 0, 0

Step 2: Imagine the body to be displaced (linear or angular as the case may be) from its mean position slightly. Find

the net restoring force (in case of linear SHM) or torque (in case of angular SHM) acting on the body in terms of displacement, ending up finally in the form of

F = kx (for linear SHM)


or = k (for angular SHM) where k should be a constant.

Step 3: Dividing F by M or ( by I, as the case may be) find the net acceleration as :

xdt

xd 22

2

or 22

2

dt

d

Knowing the value of , time period can be calculated as

T = 2

2. ENERGY METHOD: One can also find the total energy of the system, in step 2 and differentiate the expression with respect to time t and equate it to zero. [ The T.E. in SHM remains constant]. One should then eventually end up in equation of the form as obtained in step 3.

Illustration 36 : A solid cylinder is attached to a horizontal massless spring so that it can roll

without slipping along a horizontal surface. The spring constant for the spring is k. If the system is released from rest, by slightly stretching it, show that the centre of the cylinder executes Linear SHM and find its frequency of oscillations. (Figure)

Sol. Step 1: At equilibrium, no force acts horizontally, so, as such no equilibrium condition exists. Step 2: Let the system be, at any instant, such that, the spring is stretched by an amount x. If v be linear

velocity of the centre of mass of the cylinder then total energy of the system will be E = translational K.E. of cylinder + rotational K.E. of cylinder + elastic P.E. in the spring

= 2

1Mv2 +

2

1I2 +

2

1kx2

2

1Mv2 + 2

2

22

kx2

1

R

v

2

MR

2

1

[ Moment of inertia I for the cylinder about the axis of rotation is 2

MR 2

and for pure rolling motion =R

v]

Now, since no work is done against friction in pure rolling motion, so T.E. remains conserved.

or 0dt

dE

0kx2

1Mv

4

3

dt

d 22

0xM3

k2

dt

xd0

dt

kxdx

dt

dv

2

Mv32

2

Comparing the above equation with 2

2

dt

xd + 2x, it is evident that, the cylinder executes SHM, with an angular

frequency =M3

k2

Frequency of oscillations n = M3

k2

2

1

2


COMPOSITION OF TWO SIMPLE HARMONIC MOTIONS (A) Composition of two Simple Harmonic motions in same direction: Suppose two forces act on a particle, the first alone would produce a simple harmonic motion given by x1 = A1 tsin and the second aloe would produce a simple harmonic motion given by x2 = A2 sin )t( .

Both the motions are along the x-direction. The amplitudes may be different and their phases differ by . Their frequency is assumed to be same. The resultant position of the particle is then given by x = x1 + x2 = )tsin(AtsinA 21

= sintcosAcostsinAtsinA 221

= tcos)sinA(tsin)cosAA( 221

= tsinDtsinC

=

tcos

DC

Dtsin

DC

CDC

2222

22 …….. (i)

where C = A1 + A2 cos and D = A2 sin .

Now22 DC

C

and

22 DC

D

both have magnitudes less than 1 and the sum of their squares is 1. Thus, we can find

an angle between 0 and 2 such that

2222 DC

Ccosand

DC

Dsin

Equation (i) then becomes

x = )tcossintsin(cosDC 22

or x = )tsin(A ........... (A) where

A = 22 DC

= 22

221 )sinA()cosAA(

= 222

22222

21 sinAcosAcosAA2A(

= 22211 AcosAA2A ……….. (B)

and

cosAA

sinA

C

Dtan

21

2 ……… (C)

Equation (A) shows that the resultant of two simple harmonic motions along the same direction is, itself a simple harmonic motion. The amplitude and phase of the resultant simple harmonic motion depend on the amplitude of the two component simple harmonic motions as well as the phase difference between them. Amplitude of The Resultant Simple Harmonic Motion The amplitude of the resultant simple harmonic motion is given by equation (B),

2221

21 AcosAA2AA

If 0 , the two simple harmonic motions are in phase and

2221

21 AcosAA2AA = A1 A2 or A2 A1.

As the amplitude is always positive we can write A = |A1 A2|. If A1 = A2, the resultant amplitude is zero and the particle does not oscillate at all. For any value of other than O and the resultant amplitude is between |A1 A2| and A1 + A2.


A

2A

1A

Vector Method of Combining Two Simple Harmonic Motions There is a very useful method to remember the equations of resultant simple harmonic motion when two simple harmonic motion by a vector of magnitude A1 and the second simple harmonic motion by another vector of magnitude A2. We draw these vectors in figure. The vector A2 is drawn at an angle with A1 to represent that the second simple harmonic motion has a phase difference of with the first simple harmonic motion.

The resultant A

of these two vectors will represent the resultant simple harmonic motion. As we know from vectors algebra, the magnitude of the resultant vector is

Which is same an equation (12.27). The resultant A

akes an angle with 1A

where

cosAA

sinAtan

21

2

Which is same as equation (12.28). This method can easily be easily be extended to more then tow vectors. Figure shows the construction for adding three simple harmonic motions in the same direction. x1 = A1 tsin x2 = A2 )tsin( 1

x3 = A3 )tsin( 2 . The resultant motion is given by x =A sin(wt+&). (B) Composition of Two Simple Harmonic Motions in Perpendicular Directions Suppose two forces act on a particle, the first alone would produce a simple harmonic motion in xdirection give by x = tsinA1 ….(i) And the second would produce a simple harmonic motion in ydirection by given by y = )tsin(A2 ….. ( ii) The amplitudes A1 and A2 may be different and their phase differ by . The frequencies of the two simple harmonic motions are assumed to be equal. The resultant motion of the particle is a combination of the two simple harmonic motions. The position of the particle at time t is (x, y) where x is given by equation (i) and y is given by (ii) The motion is thus two-dimensional and the path of the particle is in general an ellipse. the equation of the path may be obtained by eliminating t from (i) and (ii). By (i),

1A

xtsin

Thus, 21

2

A

x1tcos

Putting in (ii), ]sintcoscost[sinAy 2

sin

A

x1cos

A

xA

21

2

12

or,

2

21

22

12sin

A

x1cos

A

x

A

y

or, 221

2

2122

2cos

A

xcos

AA

xy2

A

y

221

22 sin

A

xsin

A

2A

3A

1A 1 2


or

2

2122

2

21

2sin

AA

cosxy2

A

y

A

x ……… (D)

This is an equation of an ellipse and hence the particle moves in ellipse. Equation (i) shows that x remains between A1 and +A1 and (ii) shows that y remains between A2 and A2.Thus, the particle always remains inside the rectangle defined by x = + A1 , y = + A2. The ellipse given by (D) is traced inside this rectangle and touches it on all the four sides (figure) Special Cases

(a) 0 The two simply harmonic motions are in phase. When the x-coordinate of the particle crosses the value 0, the y-coordinate also crosses the value 0. When x-coordinate reaches its maximum value A1, the y-coordinate also reaches its maximum value A2. Similarly when x-coordinate reaches its minimum value A1, the y-coordinate reaches its minimum value A2. If we substitute = 0 in equation (D), we get

0AA

xy2

A

x

A

x

2122

2

21

2

or 0A

y

A

x2

21

or y xA

A

1

2 ………. (iii)

Which is the equation of a straight line passing through the origin and having a slope

1

21

A

Atan . Figure show the path. Equation (iii) represents the diagonal AC of the

rectangle. The particle moves on this diagonal. Equation (iii) can be directly obtained by dividing (i) by (ii) and putting 0 . The displacement of the particle on this straight line at time t is

22

21

22 )tsinA()tsinA(yxr

tsin)AA( 22

21

Thus, the resultant motion is a simple harmonic motion with same frequency and phase as the component motions.

The amplitude of the resultant simple harmonic motion is 22

21 AA as is also clear from figure.

0AA

xy2

A

x

A

x

2122

2

21

2

or 0A

y

A

x2

21

or y xA

A

1

2

which is the equation of the line BD in figure Thus, the particle oscillates on the diagonal BD of the rectangle as shown in figure. The displacement on this line at time t may be obtained from equation (i) and (ii) (with ).

22

21

22 )]tsin(A[]tsinA[yxr

D CY

A B

O X2A2

1A2

D CY

A B

OX

D CY

A B

OX


tsinAAtsinAtsinA 22

21

222

221

Thus, the resultant motion is a simple harmonic motion with amplitude 22

21 AA .

(c) 2/ The two dimple harmonic motions differ in phase by 2/ . Equations (i) and (ii) may be written as x = A1 sin t y = A2 tcosA)2/tsin( 2 . Putting 2/ in equation (D), we get

1A

y

A

x22

2

21

2

which is the standard equation of an ellipse with its axes along X and Y-axes and with its centre at the origin. The length of the major and minor axes are 2A1 and 2A2. If A1 = A2 = A together with 2/ , the rectangle of figure becomes a square and the ellipse becomes a circle. Equation (D) becomes x2 + y2 = A2 which represents a circle. Thus, the combination of two simple harmonic motions of equal amplitude in perpendicular directions differing in phase by 2/ is a circular motion. The circular motion may be clockwise or anticlockwise depending on which component leads the other. Illustration 37: Find the amplitude of the simple harmonic motion obtained by combining the motions

1x 2.0cm sin t

and 2x 2.0cm sin t / 3 .

Sol: The two equations given represent simple harmonic motions along X-axis with amplitudes A1 = 2.0 cm and A2 = 2.0 cm. The phase difference between the two simple harmonic motions is / 3 .The resultant simple harmonic motion will have an amplitude A given by

2 21 2 1 2A A A 2A A cos

= 2 2 22.0cm 2.0cm 2 2.0cm cos

3

= 3.5 cm Illustration 38: A Particle is subjected to two simple harmonic motions 1 1x A sin t

and 2 2x A sin t / 3 .

Find (a) the displacement at t = 0, (b) the maximum speed of the particle and (c) the maximum acceleration of the particle. Sol: (a) At t = 0, 1 1x A sin t 0

and 2 2x A sin t / 3 = 22

A 3A sin / 3

2

Thus, the resultant displacement at t = 0 is

21 2

A 3x x x

2 .

(b) The resultant of the two motions is a simple harmonic motion of the same angular frequency . The amplitude of the resultant motion is

2 21 2 1 2A A A 2A A cos / 3

= 2 21 2 1 2A A A A

The maximum speed is

D CY

A BH

X

F

EC


2 2max 1 2 1 2u A A A A A

(c) The maximum acceleration is

2 2 2 2max 1 2 1 2a A A A A A

Illustration 39: A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motions, find the phase difference between the individual motions. Sol: Let the amplitudes of the individual motions be A each. The resultant amplitude is also A. If the phase difference between the two motion is .

2 2A A A 2A.A cos

Or = A 2 1 cos A cos2

Or, 1

cos2 2

Or, 2 / 3.

INTRODUCTION SIMPLE HARMONIC MOTION LINEAR SIMPLE ...

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