Simple Harmonic Motion

Reading: Chapter 15

Simple Harmonic Motion

Frequency f

Period T

Tf

1

.

Simple harmonic motion

).cos()( txtx m

Amplitude xm

Phase

Angular frequency

Since the motion returns to its initial value after one

period T,

)],)(cos[)cos( Ttxtx mm

t t T 2 ( ) ,

T 2 .

Thus

2

2T

f .

Velocity

)],cos([)( txdt

d

dt

dxtv m

).sin()( txtv m

Velocity amplitude v xm m .

Acceleration

)],sin([)( txdt

d

dt

dvta m

).cos()( 2 txta m

Acceleration amplitude a xm m2 .

Note that

),()( 2 txta

.02

2

2

xdt

xd

This equation of motion will be

very useful in identifying

simple harmonic motion and its

frequency.

The Force Law for Simple Harmonic Motion

Consider the simple harmonic motion of a block of mass

m subject to the elastic force of a spring. Newton’s law:

F kx ma .

.02

2

kxdt

xdm

.02

2

xm

k

dt

xd

Comparing with the equation of motion for simple

harmonic motion,

.2

m

k

Simple harmonic motion is the motion executed by a

particle of mass m subject to a force that is proportional

to the displacement of the particle but opposite in sign.

Angular frequency:

k

m.

Period: Since T 2

,

Tm

k 2 .

Examples

15-1 A block whose mass m is 680 g is fastened to a

spring whose spring constant k is 65 Nm-1. The block is

pulled a distance x = 11 cm from its equilibrium position

at x = 0 on a frictionless surface and released from rest at

t = 0.

(a) What are the angular frequency, the frequency, and

the period of the resulting oscillation?

(b) What is the amplitude of the oscillation?

(c) What is the maximum speed of the oscillating block?

(d) What is the magnitude of the maximum acceleration

of the block?

(e) What is the phase constant for the motion?

(f) What is the displacement function x(t)?

(a) 1rads 78.968.0

65 m

k (ans)

Hz 56.12

f (ans)

s 643.01

fT (ans)

(b) cm 11mx (ans)

(c) 1ms 08.1)11.0)(78.9( mm xv (ans)

(d) 222 ms 5.10)11.0()78.9( mm xa (ans)

(e) At t = 0,

11.0cos)0( mxx (1)

0sin)0( mxv (2)

(2): 0sin 0 (ans)

(f) )78.9cos(11.0)cos()( ttxtx m (ans)

15-2 At t = 0, the displacement of x(0) of the block in a

linear oscillator is 8.50 cm. Its velocity v(0) then is

0.920 ms1, and its acceleration a(0) is +47.0 ms2.

(a) What are the angular frequency ?

(b) What is the phase constant and amplitude xm?

(a) )cos()( txtx m

)sin()( txtv m

)cos()( 2 txta m

At t = 0,

085.0cos)0( mxx (1)

920.0sin)0( mxv (2)

0.47cos)0( 2 mxa (3)

(3) (1): 2

)0(

)0(

x

a

1rads 5.230850.0

0.47

)0(

)0(

x

a (ans)

(c) (2) (1):

tan

cos

sin

)0(

)0(

x

v

4603.0)085.0)(51.23(

920.0

)0(

)0(tan

x

v

o7.24 or ooo 1557.24180

(1): cos

)0(xxm

If = 24.7o, cm 4.9m 094.07.24cos

085.0o

mx

If = 155o, cm 4.9m 094.0155cos

085.0o

mx

Since xm is positive, = 155o and xm = 9.4 cm. (ans)

Energy in Simple Harmonic Motion

Potential energy:

Since ),cos()( txtx m

).(cos2

1

2

1)( 222 tkxkxtU m

Kinetic energy:

Since ),sin()( txtv m

).(sin2

1

2

1)( 2222 txmmvtK m

Since 2 = k/m,

).(sin2

1)( 22 tkxtK m

Mechanical energy:

)(sin2

1)(cos

2

1 2222 tkxtkxKUE mm

)].(sin)([cos2

1 222 ttkxm

Since cos2(t + ) + sin2(t + ) = 1,

.2

1 2

mkxKUE

The mechanical energy is conserved.

15-3 Suppose the damper of a tall building has mass m =

2.72 105 kg and is designed to oscillate at frequency f =

10 Hz and with amplitude xm = 20 cm.

(a) What is the total mechanical energy E of the damper?

(b) What is the speed of the damper when it passes

through the equilibrium point?

See Youtube “Discovery Channel Taipei 101 (3/5)” and

“Taipei 101 Damper”

(a) 22 )2( fmmk

N 10073.1)20)(1072.2( 925

The energy:

22

2

1

2

1kxmvUKE

29 )2.0)(10073.1(2

10

MJ 521J 10147.2 7 . (ans)

(b) Using the conservation of energy,

22

2

1

2

1kxmvUKE

0)1072.2(2

110147.2 257 v

1ms 6.12 v (ans)

An Angular Simple Harmonic Oscillator

When the suspension wire is twisted through an angle ,

the torsional pendulum produces a restoring torque

given by

.

is called the torsion constant.

Using Newton’s law for angular motion, I ,

I , d

dt I

2

20

.

Comparing with the equation of motion for simple

harmonic motion,

2 I

.

Period: Since T 2

,

TI

2

.

Example

15-4 A thin rod whose length L is 12.4 cm and whose

mass m is 135 g is suspended at its midpoint from a long

wire. Its period Ta of angular SHM is measured to be

2.53 s. An irregularly shaped object, which we call X, is

then hung from the same wire, and its period Tb is found

to be 4.76 s. What is the rotational inertia of object X

about its suspension axis?

Rotational inertia of

the rod about the

center

2

12

1MLIa

2)124.0)(135.0(12

1

24 kgm 107298.1

Since

aa

IT 2 and

b

b

IT 2 , we have

b

a

b

a

I

I

T

T

Therefore,

a

a

bb I

T

TI

2

244

2

kgm 1012.6)1073.1(53.2

76.4

(ans)

The Simple Pendulum

The restoring torque about the

point of suspension is = mg

sin L.

Using Newton’s law for angular

motion, = I,

,sin 2 mLLmg

.0sin2

2

L

g

dt

d

When the pendulum swings

through a small angle, sin .

Therefore

.02

2

L

g

dt

d

Comparing with the equation of

motion for simple harmonic

motion,

.2

L

g

Period: Since T 2

,

TL

g 2 .

The Physical Pendulum

The restoring torque about the point of suspension is =

mg sin h.

Using Newton’s law for angular motion, = I,

mg h Isin , .0sin2

2

I

mgh

dt

d

When the pendulum swings through a small angle, sin

. Therefore

.02

2

I

mgh

dt

d

Comparing with the equation of motion for simple

harmonic motion,

.2

I

mgh

Period: Since T 2

,

TI

mgh 2 .

If the mass is concentrated at the center of mass C, such

as in the simple pendulum, then

.2222

g

L

mgL

mL

mgh

IT

We recover the result for the simple pendulum.

Examples

15-5 A meter stick, suspended from one end, swings as a

physical pendulum.

(a) What is its period of oscillation T?

(b) A simple pendulum oscillates with the same period as

the stick. What is the length L0 of the simple pendulum?

Rotational inertia of a

rod about one end

2

3

1ML

Period mgh

IT 2

2/

3/2

2

mgL

mL

g

L

3

22 (ans)

(b) For a simple pendulum of length L0,

g

LT 02

g

L

g

L

3

222 0 cm 7.66

3

20 LL (ans)

15-6 A diver steps on the diving board and makes it

move downwards. As the board rebounds back through

the horizontal, she leaps upward and lands on the free

end just as the board has completed 2.5 oscillations

during the leap. (With such timing, the diver lands when

the free end is moving downward with greatest speed.

The landing then drives the free end down substantially,

and the rebound catapults the diver high into the air.)

Modeling the spring board as the rod-spring system (Fig.

15-12(d)), what is the required spring constant k? Given

m = 20 kg, diver’s leaping time tfl = 0.62 s.

See Youtube “Guo Jingjing”.

Damped Simple Harmonic Motion

The liquid exerts a damping force

proportional to the velocity. Then,

,bvFd

b = damping constant.

Using Newton’s second law,

.makxbv

.02

2

kxdt

dxb

dt

xdm

Solution:

),'cos()( 2/ textx mbt

m where

.4

'2

2

m

b

m

k

If b = 0, ’ reduces to mk / of the undamped

oscillator. If kmb , then ’ .

The amplitude, mbt

mextx 2/)( , gradually decreases with

time.

The mechanical energy decreases exponentially with

time.

.2

1)( /2 mbt

mekxtE

Example

15-7 For the damped oscillator with m = 250 g, k = 85

Nm1, and b = 70 gs1.

(a) What is the period of the motion?

(b) How long does it take for the amplitude of the

damped oscillations to drop to half its initial value?

(c) How long foes it take for the mechanical energy to

drop to half its initial value?

(a) s 34.085

25.022

k

mT (ans)

(b) When the amplitude drops by half,

m

mbt

m xex2

12/

2

12/ mbte

Taking logarithm, 2ln2

1ln

2

m

bt

s 95.407.0

)2)(ln25.0)(2(2ln2

b

mt (ans)

(c) When the energy drops by half,

2/2

2

1

2

1

2

1m

mbt

m kxekx

2

1/ mbte

Taking logarithm, 2ln2

1ln

m

bt

s 48.207.0

)2)(ln25.0(2ln

b

mt (ans)

Forced Oscillations and Resonance

When a simple harmonic oscillator is driven by a

periodic external force, we have forced oscillations or

driven oscillations.

Its behavior is determined by two angular frequencies:

(1) the natural angular frequency

(2) the angular frequency d of the external driving

force.

The motion of the forced oscillator is given by

x t x t

m d( ) ). cos(

(1) It oscillates at the angular frequency d of the

external driving force.

(2) Its amplitude xm is greatest when

d .

This is called resonance.

See Youtube “Tacoma Bridge Disaster”.