Chapter 14 - Simple Harmonic Motion A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007.

Chapter 14 - Simple Chapter 14 - Simple Harmonic MotionHarmonic Motion

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

Photo by Mark Tippens

A TRAMPOLINE exerts a restoring force on the jumper that is directly proportional to the average force required to displace the mat. Such restoring forces provide the driving forces necessary for objects that oscillate with simple harmonic motion.

Objectives: After finishing Objectives: After finishing this unit, you should be this unit, you should be able to:able to:

• Write and apply Write and apply Hooke’s LawHooke’s Law for objects for objects moving with simple harmonic motion.moving with simple harmonic motion.

• Describe the motion of Describe the motion of pendulumspendulums and calculate the and calculate the length length required to produce a required to produce a given given frequency.frequency.

• Write and apply formulas for Write and apply formulas for finding the finding the frequency frequency ff,, period period TT,, velocity velocity vv, or , or accelerationacceleration aa in terms of in terms of displacementdisplacement xx or or time time tt..

Periodic MotionPeriodic MotionSimple periodic motionSimple periodic motion is that motion in is that motion in which a body moves back and forth over a which a body moves back and forth over a fixed path, returning to each position and fixed path, returning to each position and velocity after a definite interval of time.velocity after a definite interval of time.

AmplitudeA

PeriodPeriod, T, is the time for one complete oscillation. (seconds,s)(seconds,s)

PeriodPeriod, T, is the time for one complete oscillation. (seconds,s)(seconds,s)

FrequencyFrequency, f, is the number of complete oscillations per second. Hertz (sHertz (s-1-1))

FrequencyFrequency, f, is the number of complete oscillations per second. Hertz (sHertz (s-1-1))

1f

T

Example 1:Example 1: The suspended mass The suspended mass makes 30 complete oscillations in 15 s. makes 30 complete oscillations in 15 s. What is the period and frequency of the What is the period and frequency of the motion?motion?

x FF

15 s0.50 s

30 cylcesT

Period: T = 0.500 sPeriod: T = 0.500 s

1 1

0.500 sf

T Frequency: f = 2.00 HzFrequency: f = 2.00 Hz

Simple Harmonic Motion, Simple Harmonic Motion, SHMSHM

Simple harmonic motionSimple harmonic motion is periodic motion in is periodic motion in the absence of friction and produced by a the absence of friction and produced by a restoring force that is directly proportional to restoring force that is directly proportional to the displacement and oppositely directed.the displacement and oppositely directed.

A restoring force, F, acts in the direction opposite the displacement of the oscillating body.

F = -kx

A restoring force, F, acts in the direction opposite the displacement of the oscillating body.

F = -kx

x FF

Hooke’s LawHooke’s LawWhen a spring is stretched, there is a

restoring force that is proportional to the displacement.

F = -kx

The spring constant k is a property of the spring given

by:

k = F

x

F

x

m

Work Done in Stretching a Work Done in Stretching a SpringSpring

F

x

m

Work done Work done ONON the spring is the spring is positivepositive; work ; work BYBY spring is spring is

negative.negative.From Hooke’s law the force F From Hooke’s law the force F

is:is:

F (x) = kx

x1 x2

F

To stretch spring To stretch spring from xfrom x11 to x to x2 2 , work , work

is:is:2 22 1½ ½Work kx kx 2 2

2 1½ ½Work kx kx

(Review module on work)(Review module on work)

Example 2:Example 2: A 4-kg mass suspended A 4-kg mass suspended from a spring produces a displacement from a spring produces a displacement of 20 cm. What is the spring constant?of 20 cm. What is the spring constant?

F20 cm

m

The stretching force is the The stretching force is the weight (W = mg) of the 4-kg weight (W = mg) of the 4-kg

mass:mass:

F = F = (4 kg)(9.8 m/s(4 kg)(9.8 m/s22) = 39.2 N) = 39.2 N

Now, from Hooke’s law, the force Now, from Hooke’s law, the force constant k of the spring is:constant k of the spring is:

k = =k = =

FF

xx

0.2 m0.2 mk = 196

N/mk = 196

N/m

Example 2(cont.:Example 2(cont.: The mass The mass m m is now is now stretched a distance of 8 cm and held. stretched a distance of 8 cm and held. What is the What is the potential energypotential energy? (k = 196 ? (k = 196 N/m)N/m)

F8 cm

m

U = 0.627 JU = 0.627 J

The potential energy is The potential energy is equal to the work done in equal to the work done in stretching the spring:stretching the spring:

2 22 1½ ½Work kx kx

0

2 2½ ½(196 N/m)(0.08 m)U kx

Displacement in SHMDisplacement in SHM

m

x = 0 x = +Ax = -A

x

• Displacement is positive when the position is to the right of the equilibrium position (x = 0) and negative when located to the left.

• The maximum displacement is called the amplitude A.

Velocity in SHMVelocity in SHM

m

x = 0x = 0 x = +Ax = +Ax = -Ax = -A

v (+)v (+)

• Velocity is Velocity is positive positive when moving to the when moving to the rightright and negative when moving to the and negative when moving to the left.left.

• It is It is zerozero at the end points and a at the end points and a maximummaximum at the midpoint in either at the midpoint in either direction (+ or -).direction (+ or -).

v (-)v (-)

Acceleration in SHMAcceleration in SHM

m

x = 0 x = +Ax = -A• Acceleration is in the direction of the Acceleration is in the direction of the

restoring forcerestoring force. (. (aa is is positive positive when when xx is negative, and is negative, and negativenegative when x is when x is positive.)positive.)

• Acceleration is a maximum at the end points and it is zero at the center of oscillation.

+x-a

-x+a

F ma kx F ma kx

Acceleration vs. Acceleration vs. DisplacementDisplacement

m

x = 0 x = +Ax = -A

x va

Given the spring constant, the Given the spring constant, the displacement, and the mass, the displacement, and the mass, the accelerationacceleration can be found from: can be found from:

oror

Note: Acceleration is always Note: Acceleration is always opposite opposite to displacement.to displacement.

F ma kx F ma kx kxa

m

kxa

m

Example 3:Example 3: A A 2-kg2-kg mass hangs at the mass hangs at the end of a spring whose constant is end of a spring whose constant is k = k = 400 N/m400 N/m. The mass is displaced a . The mass is displaced a distance of distance of 12 cm12 cm and released. What and released. What is the acceleration at the instant the is the acceleration at the instant the displacement is displacement is x = +7 cmx = +7 cm??

m+x

(400 N/m)(+0.07 m)

2 kga

a = -14.0 m/s2a = -14.0 m/s2 a

Note: When the displacement is Note: When the displacement is +7 cm+7 cm (downward), the acceleration is (downward), the acceleration is -14.0 -14.0 m/sm/s22 (upward) independent of motion (upward) independent of motion direction.direction.

kxa

m

kxa

m

Example 4:Example 4: What is the What is the maximum maximum acceleration for the acceleration for the 2-kg2-kg mass in the mass in the previous problem? (previous problem? (A = 12 cmA = 12 cm, , k = 400 k = 400 N/mN/m))

m+x

The maximum acceleration The maximum acceleration occurs when the restoring force occurs when the restoring force is a maximum; i.e., when the is a maximum; i.e., when the stretch or compression of the stretch or compression of the spring is largest.spring is largest.

F = ma = -kx xmax = A

400 N( 0.12 m)

2 kg

kAa

m

amax = ± 24.0 m/s2amax = ± 24.0 m/s2Maximum Maximum Acceleration:Acceleration:

Conservation of EnergyConservation of EnergyThe The total mechanical energytotal mechanical energy (U + K)(U + K) of of a vibrating system is constant; i.e., it is a vibrating system is constant; i.e., it is the same at any point in the oscillating the same at any point in the oscillating path.path.

m

x = 0 x = +Ax = -A

x va

For any two points A and B, we may For any two points A and B, we may write:write:

½mvA2 + ½kxA 2 = ½mvB

2 + ½kxB 2 ½mvA2 + ½kxA 2 = ½mvB

2 + ½kxB 2

Energy of a Vibrating Energy of a Vibrating System:System:

m

x = 0 x = +Ax = -A

x va

• At any other point: At any other point: U + K = ½mvU + K = ½mv22 + ½kx + ½kx22

U + K = ½kAU + K = ½kA22 x = x = A and v = 0. A and v = 0.

• At points At points AA and and BB, the velocity is zero and the , the velocity is zero and the acceleration is a maximum. The total energy is:acceleration is a maximum. The total energy is:

A B

Velocity as Function of Velocity as Function of Position.Position.

m

x = 0 x = +Ax = -A

x va

kv A

m

vmax when x = 0:

2 2kv A x

m 2 2 21 1 1

2 2 2mv kx kA

Example 5:Example 5: A A 2-kg2-kg mass hangs at the mass hangs at the end of a spring whose constant is end of a spring whose constant is k = 800 k = 800 N/mN/m. The mass is displaced a distance of . The mass is displaced a distance of 10 cm10 cm and released. What is the velocity and released. What is the velocity at the instant the displacement is at the instant the displacement is x = +6 x = +6 cmcm??

m+x

½½mvmv22 + ½kx + ½kx 22 = ½kA = ½kA22

2 2kv A x

m

2 2800 N/m(0.1 m) (0.06 m)

2 kgv

v = ±1.60 m/sv = ±1.60 m/s

Example 5 (Cont.):Example 5 (Cont.): What is the What is the maximum velocity for the previous maximum velocity for the previous problem? (problem? (A = 10 cm, k = 800 N/m, m = A = 10 cm, k = 800 N/m, m = 2 kg2 kg.).)

m+x

½mv2 + ½kx 2 = ½kA2

800 N/m(0.1 m)

2 kg

kv A

m

v = ± 2.00 m/sv = ± 2.00 m/s

0

The velocity is maximum when x = The velocity is maximum when x = 0:0:

The The reference circlereference circle compares compares the circular motion of an object the circular motion of an object with its horizontal projection.with its horizontal projection.

2f

The Reference CircleThe Reference Circle

cos(2 )x A ft cos(2 )x A ft

cosx A t

x = Horizontal x = Horizontal displacement.displacement.A = Amplitude A = Amplitude (x(xmaxmax).).

= Reference angle.= Reference angle.

Velocity in SHMVelocity in SHMThe The velocityvelocity (v) of an (v) of an oscillating body at any oscillating body at any instant is the instant is the horizontal component horizontal component of its tangential of its tangential velocity (vvelocity (vTT).).

vT = R = A; 2f

v = -vT sin ; = t

v = - A sin t

v = -2f A sin 2ftv = -2f A sin 2ft

The The accelerationacceleration ((aa)) of an of an oscillating body at any oscillating body at any instant is the horizontal instant is the horizontal component of its component of its centripetal centripetal acceleration (acceleration (aacc).).

Acceleration Reference Acceleration Reference CircleCircle

a = -ac cos = -ac cos(t)2 2 2

2; c c

v Ra a R

R R

R = A

a = -cos(t)

2 24 cos(2 )a f A ft

2 24a f x

The Period and Frequency as a The Period and Frequency as a Function of Function of aa and and xx..

For any body undergoing For any body undergoing simple harmonic simple harmonic motionmotion::

Since a = -4f2x and T = 1/f

1

2

af

x

1

2

af

x

2x

Ta

2

xT

a

The frequency and the period can be found if the displacement and acceleration are known. Note that the signs of a and x will always be opposite.

The frequency and the period can be found if the displacement and acceleration are known. Note that the signs of a and x will always be opposite.

Period and Frequency as a Period and Frequency as a Function of Mass and Spring Function of Mass and Spring

Constant.Constant.For a vibrating body with an For a vibrating body with an elastic restoring elastic restoring force:force:

Recall that Recall that F = ma = -kxF = ma = -kx:

1

2

kf

m

1

2

kf

m 2

mT

k2

mT

k

The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units.

The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units.

Example 6:Example 6: The frictionless system The frictionless system shown below has a shown below has a 2-kg2-kg mass attached mass attached to a spring (to a spring (k = 400 N/mk = 400 N/m). The mass is ). The mass is displaced a distance of displaced a distance of 20 cm20 cm to the to the right and released.right and released.What is the frequency of the motion?What is the frequency of the motion?

m

x = 0 x = +0.2 m

x va

x = -0.2 m

1 1 400 N/m

2 2 2 kg

kf

m

f = 2.25 Hzf = 2.25 Hz

Example 6 (Cont.):Example 6 (Cont.): Suppose the Suppose the 2-kg2-kg mass of the previous problem is mass of the previous problem is displaced displaced 20 cm20 cm and released ( and released (k = 400 k = 400 N/mN/m). What is the maximum ). What is the maximum acceleration? (acceleration? (f = f = 2.25 Hz2.25 Hz))

m

x = 0 x = +0.2 m

x va

x = -0.2 m

2 2 2 24 4 (2.25 Hz) ( 0.2 m)a f x

Acceleration is a maximum when Acceleration is a maximum when x = x = A A

a = 40 m/s2a = 40 m/s2

Example 6:Example 6: The The 2-kg2-kg mass of the mass of the previous example is displaced initially previous example is displaced initially at at x = 20 cmx = 20 cm and released. What is the and released. What is the velocity velocity 2.69 s2.69 s after release? (Recall after release? (Recall that that ff = 2.25 Hz = 2.25 Hz.).)

m

x = 0 x = +0.2 m

x va

x = -0.2 m

v = -0.916 m/sv = -0.916 m/s

v = -2f A sin 2ftv = -2f A sin 2ft

2 (2.25 Hz)(0.2 m)sin 2 (2.25 Hz)(2.69 s)v

(Note: in rads) 2 (2.25 Hz)(0.2 m)(0.324)v

The minus sign means The minus sign means it is moving to the left.it is moving to the left.

Example 7:Example 7: At what time will the 2-kg At what time will the 2-kg mass be located 12 cm to the left of x mass be located 12 cm to the left of x = 0? = 0? (A = 20 cm, f = 2.25 Hz)(A = 20 cm, f = 2.25 Hz)

m

x = 0 x = +0.2 m

x va

x = -0.2 m

t = 0.157 st = 0.157 s

cos(2 )x A ft

-0.12 m

10.12 mcos(2 ) ; (2 ) cos ( 0.60)

0.20 m

xft ft

A

2.214 rad2 2.214 rad;

2 (2.25 Hz)ft t

The Simple PendulumThe Simple Pendulum

The period of a The period of a simple simple pendulumpendulum is given by: is given by:

mg

L

2L

Tg

For small angles

1

2

gf

L

Example 8.Example 8. What must be the length of a simple pendulum for a clock which has a period of two seconds (tick-tock)?

2L

Tg

L

22 2

24 ; L =

4

L T gT

g

2 2

2

(2 s) (9.8 m/s )

4L

L = 0.993 m

The Torsion PendulumThe Torsion Pendulum

The period The period TT of a of a torsion pendulumtorsion pendulum is is given by:given by:

Where k’ is a torsion constant that depends on the material from which the rod is made; I is the rotational inertia of the vibrating system.

Where k’ is a torsion constant that depends on the material from which the rod is made; I is the rotational inertia of the vibrating system.

2'

IT

k

Example 9:Example 9: A A 160 g160 g solid disk is solid disk is attached to the end of a wire, then attached to the end of a wire, then twisted at twisted at 0.8 rad0.8 rad and released. The and released. The torsion constant k’ is torsion constant k’ is 0.025 N m/rad0.025 N m/rad. . Find the period.Find the period.

(Neglect the torsion in the wire)(Neglect the torsion in the wire)

For DiskFor Disk: I = ½mRI = ½mR22

I = ½(0.16 kg)(0.12 m)I = ½(0.16 kg)(0.12 m)22 == 0.00115 kg m0.00115 kg m22

20.00115 kg m2 2

' 0.025 N m/rad

IT

k T = 1.35 sT = 1.35 s

Note: Period is independent of angular displacement.Note: Period is independent of angular displacement.

SummarySummary

Simple harmonic motion (SHM)Simple harmonic motion (SHM) is that is that motion in which a body moves back and motion in which a body moves back and forth over a fixed path, returning to each forth over a fixed path, returning to each position and velocity after a definite position and velocity after a definite interval of time.interval of time.

Simple harmonic motion (SHM)Simple harmonic motion (SHM) is that is that motion in which a body moves back and motion in which a body moves back and forth over a fixed path, returning to each forth over a fixed path, returning to each position and velocity after a definite position and velocity after a definite interval of time.interval of time.

1f

TF

x

m

The frequency (rev/s) is the reciprocal of the period (time for one revolution).

The frequency (rev/s) is the reciprocal of the period (time for one revolution).

Summary (Cont.)Summary (Cont.)

F

x

m

Hooke’s Law: In a spring, there is a restoring force that is proportional to the displacement.

Hooke’s Law: In a spring, there is a restoring force that is proportional to the displacement.

The spring constant k is defined by:

Fk

x

Fk

x

F kxF kx

Summary (SHM)Summary (SHM)

F ma kx F ma kx kxa

m

kxa

m

m

x = 0 x = +Ax = -A

x va

½mvA2 + ½kxA 2 = ½mvB

2 + ½kxB 2 ½mvA2 + ½kxA 2 = ½mvB

2 + ½kxB 2

Conservation of Energy:

Summary (SHM)Summary (SHM)

2 2kv A x

m

2 2kv A x

m

2 2 21 1 12 2 2mv kx kA 2 2 21 1 1

2 2 2mv kx kA

0

kv A

m0

kv A

m

cos(2 )x A ft cos(2 )x A ft

2 sin(2 )v fA ft 2 sin(2 )v fA ft

2 24a f x 2 24a f x

Summary: Period and Summary: Period and Frequency for Vibrating Frequency for Vibrating

Spring.Spring.m

x = 0 x = +Ax = -A

x va

1

2

af

x

1

2

af

x

2x

Ta

2

xT

a

2m

Tk

2m

Tk

1

2

kf

m

1

2

kf

m

Summary: Simple Summary: Simple Pendulum and Torsion Pendulum and Torsion

PendulumPendulum

2L

Tg

1

2

gf

L

L

2'

IT

k

CONCLUSION: Chapter 14CONCLUSION: Chapter 14Simple Harmonic MotionSimple Harmonic Motion