Shm

1sincos

2sin

2cos

)sin()(cos

2

)(

iez

iiez

titAAeZ

i

i

ti

Key points

OSCILLATIONS

x

mvkxE

m

kxx

2

22

22

aget we

0dt

dEput then and

dt

dE find

2

1

2

1

,m

k-a

x,-kFmotion n Translatio(a)

System Mass Spring

p

cmpp

I

kR

mRIIxI

kRa

22

22

kxRR

aI

kxRk

p

pp

,

Motion Rotational)b(

system undampedfor good hold relations These

I

kR-get you will

zero equalit put and dt

dE find

2

1

2

1

2

1)(

p

2

222

xa

ImvkxEc cm

Oscillations

• Periodic Motion:- Any motion that repeats itself in equal intervals of time.

• Oscillatory Motion:- If a particle in periodic motion moves back and forth over the same path.

• Harmonic Motion:- The displacement of a particle in periodic motion can always be expressed in terms of sine and cosine functions therefore oscillatory motions are called Harmonic motions.

Oscillations

• The world is full of oscillatory motions– A child on a swing– A mass attached to a spring– A guitar string being played– Swinging pendulum – Atoms in molecules or in solid lattice– Air molecules as a sound wave passes by– Radio waves, microwaves and visible light are

oscillating magnetic and electric field vectors

t = 0

t = T/4

t = T/2

t = 3T/4

t = T

• Period T of a harmonic motion :

-The time required to complete one round trip of

the motion i.e. one complete oscillation

• Frequency f : Number of oscillations in 1 second

Unit :1 hertz (Hz) = one oscillation per second

= 1s-1

It is inverse of the period

Tf

1

• When the displacement of a particle at any time “t” can be given by this formula:

)cos()( txtx mIts motion is called simple harmonic motion

Simple Harmonic Motion

• The factors xm, ω and are all constants

• xm is the amplitude of the oscillation

• ω is angular frequency of the oscillation• The time varying quantity (ωt + ) is

called the phase of the motion and is called the phase constant or phase angle

)cos()( txtx m


txTtx

txtx m

timeTafter itself repeats SHM

cos then 0Let

•and cosine function repeats itself when its argument (the phase) has increased by the value 2π


frequencyangularcalledis

2)( tTt

fT

22


• We can see that the curves look identical except that one is ‘taller’ than the other

• These two curves have a different maximum displacement – or amplitude xm

Keeping ω & Φ constantvarying xm …

Simple Harmonic Motion• Keeping Φ & xm constant • Varying ω =2π/T …• let T´ = T/2

Simple Harmonic Motion• Keeping xm & ω constant

• varying phase angle

•The 2nd curve has ‘slid over’ by a constant amount relative to the 1st curve…

The Velocity of SHM• To get the velocity of the particle • differentiate the displacement function with respect

to time

)cos()( txtx m

)cos()( txdt

dtv m

The Velocity of SHM

)sin()( txtv m

timesallfor

txxtvor m2222

The Velocity of SHM• velocity is a sine function • It is T/4 period (or π/2) out of phase with the

displacement

The Acceleration of SHM

• lets differentiate once again to get the acceleration function:

)sin()(

)( txdt

d

dt

tvdta m

)cos()( 2 txta m


• We can combine our original equation for the displacement function and our equation for the acceleration function to get:

)()(

)cos()(2

2

txta

txta m


• This relationship of the acceleration being proportional but opposite in sign to the displacement is the hallmark of SHM

• Specifically, the constant of proportionality is the square of the angular frequency

)()( 2 txta

• The acceleration function is one-half period (or π radians) out of phase with the displacement and the maximum magnitude of the acceleration is ω2xm

• When the displacement is at a maximum, the acceleration is also at a maximum (but opposite in sign)

• And when the displacement and acceleration are at a maximum, the velocity is zero

• Similarly, when the displacement is zero, the velocity is at a maximum

• Does this remind you of anything you have seen previously?

The Force Law for SHM• Knowing the acceleration of a particle in SHM, lets now apply

Newton’s 2nd law to get the equation of motion of SHM

xm

maF2

kxThis force is called restoring force

k is called force constant

The Force Law for SHM• SHM is defined as when a particle of mass m

experiences a force that is proportional to the displacement of the particle, but opposite in sign

• Equation of SHM kxma 0

2

2

kxdt

xdmor

022

2

xdt

xdor Where ω is

natural frequency

Simple Harmonic Oscillator

• The block-spring system shown here is a classic linear simple harmonic oscillator where ‘linear’ means that the force F is proportional to the displacement x rather than to some other power of x

Mass-spring system• The angular frequency ω and period T are

therefore related to the spring constant k by the formulas:

m

kkm 2

k

mT 2

Problem 1

• Which of these relationships implies SHM?

a) F = -5x

b) F = -400x2

c) F = 10x

d) F = 3x2

Problem 2• At t = 0 the displacement x(0) of the block is -

8.50 cm, the velocity is v(0) = -0.920 m/s and it’s acceleration is a(0) = +47.0 m/s

• What are the values for ω, xm and ?

Solution of problem 2

)()( 2 txta

rad/s 23.5m/s 0.0850

m/s 47.0

)0(

)0(

2

x

a

mxv

x

txxtv

m

m

09358.000 2

2

2

2222

027.155

9083.0358.9

5.8cos

cos0

mxx

P 17.3To be on the verge of slipping means that the force exerted on the smaller block is :-

Mm

k

mmms xamamgf 2max &

k

gMmx sm

)(

x

m

kk

dt

xd

xkxkdt

xdm

212

2

212

2

P 17.5

21

21

22

KK

mT

m

kk

Continued …….

P 17.6

xkk

kx

21

12

mkk

kk

xkk

kk

dt

xdm

)(

21

21

21

212

2

222

2

xkdt

xdm

P 17.7

nkk

kk

xk

xkkx

xkxkkx

xx

xkxk

xxxa

1

1

2

1

2

12

222

211

2

21

2111

21

be part willeach ofconstant force

thenpartsn equal into divided isit If

244

2

1

2

1

2

12

x

length equal

kgMM

k

kx

xkxkdt

xdMb

43.4

4

4

212

2

P 17.7

L

gT

Lg

yL

g

dt

yd

gyAdt

ydAL

2

2

2

2

2

2

E 17.17gALmg

gyAdt

ydm

2

2

Energy in SHM

• We already know from theChapters 11 &12 on Work,Kinetic Energy, Potential Energy and Conservation of Energy that kinetic and potential energy get transferred back and forth as a linear oscillator moves

• Remember also that the total mechanical energy E remains constant

Energy in SHM

• We know from Chapter 12 that the potential energy stored in the spring is:

2

2

1)( kxtU

•Substituting x(t), we get:

)(cos2

1)( 22 tkxtU m

Energy in SHM The Kinetic energy of the block is:

2

2

1)( mvtK

)(sin2

1

2

1)( 222 tkxmvtK m

Using the velocity function and also making a substitution of ω2 = k/m we get:

Energy in SHM• the total mechanical energy is:

)(sin)(cos2

1

)(sin2

1)(cos

2

1

222

2222

ttkxE

tkxtkxE

m

mm

KUE

Energy in SHM

2

2

1mkx

KUE

So we finally get

The total energy of a linear oscillator is constant and independent of time

Energy in SHM

• So an oscillating system contains an element of ‘springiness’ (which stores the potential energy) and an element of inertia (which stores the kinetic energy) – even if the system is not mechanical in nature

• In an electrical system, capacitor is the element of ‘springiness’ and a choke (a coil i.e. inductor) is the element of ‘inertia’.

Energy Conservation in Oscillatory Motion

212K mv

212U kx

2 21 12 2E K U mv kx

At any time t

E = U (t) + K (t) = constant

constant0

constant2

1

2

1

222

22

xvr

kxmv

Differentiating w. r. t. t

00

0

22

22

2

xdt

xdiex

dt

dvor

dt

dxx

dt

dvv

Energy Law in SHM

SHO ofmotion ofequation get the we

zero to equating

constant E

dt

dE

Problem 3

• In the figure the block has a kinetic energy of 3 J and the spring has an elastic potential energy of 2 J when the block is at x = +2.0 cm

– (a) What is the KE when the block is at x = 0?– What are the elastic potential energies when the

block is at (b) x = -2.0 cm and (c) x = -xm?

•(a) KE = Kmax = E = 5J

Solution problem 3

(b) PE ( at x = - 2.0 cm) = PE ( at x = + 2.0 cm) = 2J

(c) PE ( x = - xm) = Umax = E = 5J

P 17.13

k

MT

xM

Ka

2

32

3

2

22

2

1,

2

1 IKEMvKE rt

22222

2

1

4

3

2

1

2

1

2

1kxMvIMvkxE

02

3

02

30

kxMa

dt

dxkx

dt

dvMv

dt

dE

P 17. 8

2

2

1

spring) of portion small of .(

x

L

Vdx

L

m

EKdK

Continued …..

L

spring

dxxL

mv

xL

vdx

L

mK

0

23

2

22

2

2

1

2

1

33m

M

kx

mM

ka

kxdt

dvmM

dt

dE

kxmvMvE

mvL

L

mv

30

2

1

6

1

2

1

632

1

222

23

3

2

Shm

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