Chapter 14Oscillations

1. Simple Harmonic Motion

According to Hooke’s law, the restoring force of the spring is proportional to the displacement (stretching or compression) of the spring,

xF kx= −where k is the force constant of the spring (spring stiffness).

The Newton’s second law yields

If the acceleration of an object is proportional to its displacement and is oppositely directed, the object will move with simple harmonic (sinusoidal) motion.

2

2

,x x

d x kx mdt

F kx ma

a x

= − =

≡ = −

The time it takes for an object to execute a complete cycle of oscillatory motion, is called the period T. The reciprocal of the period, is the frequency f,

1 .Tf =The unit of frequency is the cycle per second (cy/s) or hertz (Hz).

The general solution of the harmonic equation is( )cosx A tω δ= +

ω

δ

where A (maximum displacement) is the amplitude, is the angularfrequency (units: radians per second or 1/sec), is the phase of the motion, and is the phase constant. If we have only one oscillator, we can always choose t = 0 in such a way as to make

tω δ+0, cos .x tδ ω= =

If we have two systems with the same frequency,

( ) ( )1 1 2 2cos , cos ,x A t x A tω δ ω δ= + = +the difference is called the phase difference. When the phase difference

is 0 or an integer times the systems are said to be in phase. When the phase difference is an odd integer times the systems are out of phase.

1 2δ δ− 1 2δ δ−2 ,π

,πThe velocity in harmonic motion is

( )sin .dxx dtv A tω ω δ= = − +

The acceleration is

( )2

22 cos .xdv d x

x dt dta A tω ω δ= = = − +

Then2

xa xω= −

or2 / , / .k m k mω ω= =

1 12 2 .T

kfm

ωπ π= = =

2 ,π

The period and the frequency for a particle on a spring are

Each time t increases by T, the phase increases by and one cycle of the motion is completed.

The frequency and period of simple harmonic motion are independent of the amplitude.

Example 1. The motion is described by the equation

( ) ( )16.0 cos 3.2 / 4 .sy m t π= +iFind the amplitude, frequency, period, velocity, acceleration of the motion, and the position of the object at t = 4 s.

13.2 .sω −=/ 2 0.51f Hzω π= = 1/ 1.96 .T f s= =

The amplitude A = 6.0 m. The angular frequency is The frequency is

. The period is

The velocity is ( ) ( ) ( )( ) ( )

1 14

14

6.0 3.2 sin 3.2

19.2 / sin 3.2

dyy dtv m s s t

m s s t

π

π

− −

−

= = − ⋅ +

= ⋅ +

i

and the acceleration is ( ) ( ) ( )( ) ( )

2

2

21 14

2 14

6.0 3.2 cos 3.2

61.44 / sin 3.2 .

d yy dt

a m s s t

m s s t

π

π

− −

−

= = − ⋅ +

= ⋅ +

i

The position of the object at t = 4 s is

( ) ( )( )

16.0 cos 3.2 4 / 4

6.0 cos 1.585 0.087.sy m s

m m

π= ⋅ +

= = −i

Example 2. An object oscillates with angular frequency 3.2 rad/s. At t=0 the object is at 8.0 cm and has velocity – 12 cm/s. Find the amplitude, the phase constant, and the maximal velocity of the object.

The displacement and velocity at t=0 are

( )( )

( ) ( )1

cos cos 0.08 ,

sin

3.2 sin 0.12 /

dxx dt

x A t A m

v A t

A s m s

ω δ δ

ω ω δ

δ−

= + = =

= = − +

= − ⋅ =Therefore,

( ) ( )1

cos 0.08 ,

3.2 sin 12.0 /

A m

A s m s

δ

δ−

=

− ⋅ =or

( ) ( )1 13.2 tan 1.5 ,

tan 0.469, 0.438

s sδ

δ δ

− −− =

= − = −Then

max

0.08 / cos 0.088 ,0.088 3.2 / 0.283 /

A m mv A m s m s

δω

= == ⋅ = ⋅ =

0δ =

Graphs of displacement

Velocity

AccelerationWith

Simple harmonic motion and circular motion

( )( )

,cos cos ,

sin sin .x

tx A A t

v A A t

θ ω δθ ω δ

ω θ ω ω δ

= +

= = +

= − = − +

14. 2 Energy in simple harmonic motion

Potential energy

( )2 2 21 12 2 cos .U kx kA tω δ= = +

Kinetic energy

The total energy, using is

( )2 2 2 21 12 2 sin .xK mv k A tω ω δ= = +

2 / ,k mω =

( ) ( )2 2 2 2 21 1 12 2 2cos sin

totE U K

kA t kA t kAω δ ω δ

= +

= + + + =

Plots of displacement x, potential energy U, and kinetic energy Kas functions of time t.

12

,

av av

U K EU K E+ == =

Example 3. A 5-kg object oscillates with an amplitude of 6 cm and a period of 3 s.What are the total energy, maximum speed, and acceleration?

The frequency2 / /T k mω π= =

Therefore, the spring constant is2 2

2 22 24 4

95 21.93 /

T sk m m kg kg sπ πω= = = ⋅ =

Then the total energy2 41 1

2 2 21.39 36 10 0.0394totE kA J J−= = ⋅ ⋅ ⋅ =

The maximum speed and acceleration are

( ) ( )

2 2max 3

2 22 22 2max 3

0.06 / 0.126 / ,

0.06 / 0.263 /T

T

v A A m s m s

a A A m s m s

π π

π π

ω

ω

= = = ⋅ =

= = = ⋅ =

14.3 Some Oscillating Systems

Total force yF ky mgΣ = − +At equilibrium, the force is zero:

Displacement from equilibrium,

0

0

0 ,/

ky mgy mg k= − +=

0

0

' ,'

y y yy y y= −= +

The total force is now 'yF kyΣ = −

and Newton’s law2

2'' d y

y dtF ky mΣ = − =

leads to the same simple harmonic motion

( ) 2' cos , / .y A t k mω δ ω= + =

around a new equilibrium position y’=0:

Example 4. A block on a spring oscillates with frequency 10 Hz and an amplitude4 cm. A tiny bead is placed on top of the block. At what distance from the block equilibrium position does the bead lose the contact withthe block?

To lose the contact, the acceleration should be g directed downward,

( )( )

2

22

,

/ / 2

9.81/ 2 10 0.15

ya g y

y g g f

m m

ω

ω π

π

= = −

= − = −

= − ⋅ = −

(…you cannot catch a platform that is going down with acceleration larger than g!)

(the bead leaves the block when y is negative which is when the bead/block is abovethe equilibrium position since down was chosen as a positive direction)

cos .T mg φ=

2

2 sin .d sdt

m mg φ= −

The simple pendulum

The sum of projections of all forces (gravity mg and tension T) on the direction of the string is zero:

The projection of the force on the direction perpendicular to the string (tangent to the circle) gives the acceleration of the ball

sin , ,s Lφ φ φ≈ ≈:φ

( )

2

2

2 2

,

cos , / , 2 / .

ddt

mL mg

A t g L T L g

φ

πω

φ

φ ω δ ω π

= −

= + = = =

At small angles, and we get the same simple harmonic

equation for the angle

sinMgD φ

sinI Mgdτ α φ= = −2 2/d dtα φ=

( )

2

2 ,

cos , / ,

2 / 2 /

ddt

I MgD

A t MgD I

T I MgD

φ φ

φ ω δ ω

π ω π

= −

= + =

= =

A rigid object free to rotate about a horizontal axis (not through its center of mass!) is called a physical pendulum.

The object will oscillate when displaced from equilibrium. The torque about the axis is where D is the distance between the axis and the center of mass.

Newton’s second law for rotation

where I is the moment of inertia and is the angular acceleration. At small angles, we again get the same equation for the simple harmonic motion!

The physical pendulum

2 / .T I MgDπ=.D x=

2 2 2112 .cmI I MD ML Mx= + = +

2 2 2 2/12 /122 2 .ML Mx L xMgx gxT π π+ += =

( )

( )

2 2 2 2

22 2/12

2

2 2

/12 2 /12

2 21 1 1 12

2 21 112

0 2 ,

0 2 /12 ,

0 2 /12 2 1 ,

L xx

dT d L x d L xdx dx gx dx xg

x x

Lx x x

x L x

x L x

ππ

+

+ += = =

= − +

= − + = − −

Example 5. A uniform rod of mass M and length L is free to rotate about a horizontal axis at a distance x from the center of mass. Find the period of oscillations T. Find the distance x for which the period is a minimum.

The period is The distance D from the center of mass to the rotation axisThe moment of inertia, according to the parallel axis theorem is

Therefore,

The period is minimum when

2 2 /12x L=

14.4 Damped Oscillations

Damped motion – motion in which mechanical energy is dissipated (lost) by frictional forces. If, because of strong dissipation, the oscillator fails to complete even one cycle of oscillations and just moves towards the equilibrium, the motion is overdamped. If the damping is small and the amplitude of oscillations decreases with time very slowly, the motion is underdamped. Motion with the minimum damping for nonoscillatorymotion is critically damped (border between over- and underdamping).

underdamped

The damping force (friction) is often proportional to the velocity (linear damping),

.dF bv= −

Then Newton’s second law has the form

The solution of this equation is

2

2 .d x dxdtdt

ma m kx b≡ = − −

( )

( )

/ 20

20 0

0

cos ' ,

/ , ' 1 / 2 ,

/ .

tx A e t

m b b m

k m

τ ω δ

τ ω ω ω

ω

−= +

= = −

=

A0 is called the initial amplitude, τ is the decay time or the time constant ofthe oscillator.

The amplitude of oscillations changes with time as

2 2 /0 .tA A e τ−=

The frequency of oscillations 'ω becomes zero at the critical value of the frictioncoefficient b,

02 .cb mω=At b > bc the system is overdamped, at b < bc - underdamped.

The energy of an oscillator changes with time as

( )2 2 2 2 / /1 10 02 2 .t tE m A m A e E eτ τω ω − −= = =

Quality factor

A damped oscillator is described by its quality factor Q,

0Q ω τ=If the energy loss per cycle is small, ,E E∆

( ) ( )

( ) 0

/ 10 0 0

2 2

,tdE d Tdt dt

E TE Qcycle

E t E e t E t Eττ τ

π πτ ω τ

−

∆

∆ = ∆ = ∆ = − ∆ = −

= ≈ =

or

( )2/ , / 1.

cycleE EQ E Eπ

∆= ∆

Example 6. A good string on a musical instrument produces, when struck, sound which loses half its energy in about 5 sec. Find the decay time, quality factor, and the fractional energy loss per cycle if the sound frequency is about 400 Hz.

( )

/ 5 / 5 /1 10 0 02 2

5 / ln 1/ 2 5 / ln 25 / ln 2 7.21 .

t s sE E e E E e es s

s s

τ τ τ

τ ττ

− − −= ⇒ = ⇒ =

⇒ − = ⇒ =

⇒ = =

Then the quality factor

( )4

0

4

2 1.81 10 ,

/ 2 / 3.47 10 .cycle

Q f

E E Q

ω τ π τ

π −

= = = ×

∆ = = ×

14.5 Driven oscillations and resonance

To keep a damped oscillator going, the energy should be put (“pumped”) into the system. In this case, the oscillator is said to be driven. If the energy is put in at the same (higher, lower) rate it is being dissipated, the amplitude of oscillations remains constant (increases, decreases).

The driving force can sometimes be a simple harmonic motion (oscillate sinusoidally) with some frequency ωω

ωω

Eventually, the steady state motion is reached in which the system oscillates with constant amplitude (at constant energy) with frequency

0ωThe amplitude of driven oscillations depends on The natural frequency (or resonance frequency)is the frequency of oscillator without damping or driving force (for a spring, 0 /k mω = ).

If the frequency of the driving force 0ω ω= , the system

is at resonance. The amplitude of oscillations and the rate of the energy transfer to the system as a function of frequency is the highest at resonance.

0

1Q

ωω∆ =

(for weak damping)

Resonance for an oscillator(P is the averaged power delivered to an oscillator,ω∆ is the width of the resonance peak)

The amplitude of driven oscillations as a function of frequency is

( )0

2 2 2 2 20

F

m bA

ω ω ω− +=

Example 7. An object of mass 2.0 kg on a spring of force constant 450 N/m loses about 2% of its energy per cycle. Determine the resonance parameters of the system if it is driven by a sinusoidal force with amplitude of F0 = 1.2 N.

The natural (resonance) frequency is 1 1

0 / 450 / 2.0 15.0 .k m s sω − −= = =

Since ( )/ 2% 0.02,cycle

E E∆ = ≡ the Q-factor is

( )2/ 2 / 0.02 314.15.

cycleE EQ π π∆= = =From equation 0 0 /Q m bω τ ω= = we can easily find the damping coefficient b,

0 / 15.0 2.0 / 314.15 / 0.095 / .b m Q kg s kg sω= = ⋅ =Then the width of the resonance curve is

1 10 / 15.0 / 314.15 0.048 .Q s sω ω − −∆ = = =

The amplitude of oscillations at resonance is

( )0 0

2 2 2 2 2 00

1.20.095 15.0 0.84 .F F

bm bA m mωω ω ω ⋅− += = = =

Review of Chapter 14

Position function, velocity, acceleration:

( ) ( ) ( )2cos , sin , cos .x xx A t v A t a A tω δ ω ω δ ω ω δ= + = − + = − +

Angular frequency & period

2 2 / .f Tω π π= =Energy

212total av avE U K kA U K= + = = +

Spring: 0 /k mω =Pendulum: 0 /g Lω =Physical pendulum: 0 /MgD Iω =

Amplitude of damped oscillations:( )/ 2/ 2

0 0b m ttA A e A eτ −−= =

Frequency of damped oscillations: 20' 1 1/ 4Qω ω= −

Energy of damped oscillations : /0

tE E e τ−=The quality factor:

( )2

0 0 0 // /cycleE EQ m b πω τ ω ω ω ∆= = = ∆ =

Amplitude & phase of driven oscillations:

( ) ( )0

2 22 2 2 2 200

, tanF bmm b

A ωω ωω ω ω

δ−− +

= =