Chapter 15– OscillationsChapter 15– OscillationsI.I. Simple harmonic motion (SHM)Simple harmonic motion (SHM)

- Velocity- Velocity- Acceleration- Acceleration

II. Force law for SHM II. Force law for SHM - Simple linear harmonic oscillator - Simple linear harmonic oscillator (block attached to spring moving in 1D)(block attached to spring moving in 1D)III. Energy in SHMIII. Energy in SHM

IV. Angular harmonic oscillatorIV. Angular harmonic oscillator - Simple angular harmonic oscillator - Simple angular harmonic oscillator

(torsion pendulum)(torsion pendulum)V. PendulumV. Pendulum

VI. SHM and uniform circular motionVI. SHM and uniform circular motion

VII. Damped simple harmonic motionVII. Damped simple harmonic motion

VIII. Forced oscillations and resonanceVIII. Forced oscillations and resonance

I. Simple harmonic motionI. Simple harmonic motion

Examples:Examples: boats, guitar strings, diaphragms in telephones, boats, guitar strings, diaphragms in telephones, oscillations of air molecules that transmit the sensation of oscillations of air molecules that transmit the sensation of sound, oscillations of atoms in a solid that convey the sound, oscillations of atoms in a solid that convey the sensation of temperature. sensation of temperature.

In the “real world” the oscillations are damped In the “real world” the oscillations are damped motion dies out gradually motion dies out gradually Mechanical energy transferred to thermal (friction) Mechanical energy transferred to thermal (friction)

I. Simple harmonic motionI. Simple harmonic motion

- Frequency:- Frequency:

Tf

1 1 Hertz = 1 oscillation per second = 1s1 Hertz = 1 oscillation per second = 1s-1-1Units:Units:

Number of oscillations completed each secondNumber of oscillations completed each second.

T = oscillation period time for a complete oscillation (cycle).

- Periodic / harmonic motion:- Periodic / harmonic motion:

Any motion that repeats itself in a particular way. Any motion that repeats itself in a particular way.

)cos()( txtx mDisplacement:Displacement: xxmm, , ω, φ =constω, φ =const. .

Magnitude of the maximum displacementMagnitude of the maximum displacement..

Simple Harmonic Motion (SHM) Periodic motion sinusoidal (sine or cosine) function of time.

Amplitude (xAmplitude (xmm): ):

Phase constant (φ): Angle that depends on the particle’s Angle that depends on the particle’s displacement and velocity at t=0. displacement and velocity at t=0.

Phase of the motion:Phase of the motion: ((ωt + φωt + φ) )

- Periodic / harmonic motion:- Periodic / harmonic motion:

Any motion that repeats itself in a particular way. Any motion that repeats itself in a particular way.

)cos()( txtx m

Particle oscillating maximum speed at x=0 minimum speed at x= ± xm

Angular frequency (Angular frequency (ωω): ):

)(coscos Ttxtx mm

After one period After one period x(T+t)=x(t) x(T+t)=x(t)

22)( TtTt

The cosine function first repeat itself after 2The cosine function first repeat itself after 2π = 1 periodπ = 1 period

fT

22

rad/srad/sUnits:Units:

Velocity of SHM:Velocity of SHM:

)sin()cos()(

)( txtxdt

d

dt

tdxtv mm

Velocity amplitude (vVelocity amplitude (vmm): ): vvmm= = ω·xω·xmm

t

Acceleration of SHM:Acceleration of SHM:

)cos()sin()()(

)( 22

2

txtxdt

d

dt

tdv

dt

txdta mm

Acceleration amplitude (aAcceleration amplitude (amm): ): aamm= = ωω22·x·xmm

)()cos()( 22 txtxta m

In In SHMSHM, the acceleration is proportional to the displacement but opposite , the acceleration is proportional to the displacement but opposite in sign and the two quantities are related by the square of the angular in sign and the two quantities are related by the square of the angular frequency.frequency.

Different amplitude

Different period

Different phase angle

t

t

II. Force law for SHMII. Force law for SHM

xmmaF )( 2

Restoring force similar to: Restoring force similar to:

kxF

Spring constant:Spring constant: mmωω22

SHM 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.

Linear simple harmonic oscillatorLinear simple harmonic oscillator

Angular frequency:Angular frequency:m

k Period:Period:

k

mT

22

Kinetic energy:

)(sin2

1)(/)(sin

2

1

2

1)( 2222222 tkxtKmktxmmvtK mm

Mechanical energy:Mechanical energy:

2

2222222

2

1

)](cos)([sin2

1)(sin

2

1)(cos

2

1

m

mmm

kxKUE

ttkxtkxtkxKUE

III. Energy in SHMIII. Energy in SHM

Linear oscillator Linear oscillator ΔΔEEmecmec= cte = cte back and forth back and forth

energy transfer between kinetic and potential energy transfer between kinetic and potential energy.energy.

)(cos2

1

2

1)( 222 tkxkxtU m

Potential energy:Potential energy:

Mechanical energy:Mechanical energy:

2

2222222

2

1

)](cos)([sin2

1)(sin

2

1)(cos

2

1

m

mmm

kxKUE

ttkxtkxtkxKUE

IV. Angular Simple Harmonic oscillatorIV. Angular Simple Harmonic oscillator

Torsion pendulumTorsion pendulumIf we rotate the disk some angular displacement If we rotate the disk some angular displacement θθ from its rest position it will oscillate about that from its rest position it will oscillate about that position in angular simple harmonic motion.position in angular simple harmonic motion.

Restoring torque:Restoring torque:

Torsion constant:Torsion constant: κκ (Greek letter (Greek letter kappa)kappa)(depends on length, diameter and material (depends on length, diameter and material of suspension wire)of suspension wire)

Angular form of Hooke’s lawAngular form of Hooke’s law

I

T 2

Angular simple harmonic oscillatorAngular simple harmonic oscillator

Period:Period:

Angular simple harmonic motion

Angular acceleration: Angular acceleration: α α = = - ω- ω22 θ θ

Linear simple harmonic motionLinear simple harmonic motion

Linear acceleration: Linear acceleration: a a = = -ω-ω22 x x

Angular acceleration proportional to angular displacement from the Angular acceleration proportional to angular displacement from the equilibrium position but tends to rotate the system in the opposite equilibrium position but tends to rotate the system in the opposite direction to the displacement.direction to the displacement.

Torque:Torque:Force: Force: F F = = -k x-k x

Torque is proportional to the angular displacement but tends to rotate Torque is proportional to the angular displacement but tends to rotate the system in the opposite direction. the system in the opposite direction.

V. PendulumV. Pendulum

Simple pendulumSimple pendulum

Particle of mass Particle of mass “m”“m” (pendulum’s (pendulum’s bob) suspended from one end of bob) suspended from one end of an unstretchable massless string an unstretchable massless string of length of length “L”“L” that is fixed at the that is fixed at the other end.other end.

Restoring torque about pendulum’s pivot point:Restoring torque about pendulum’s pivot point:

ImgLFLFr g )sin()sin(

Equilibrium positionEquilibrium position: : θ = 0θ = 0 ““I” is the pendulum’s rotational inertia about the I” is the pendulum’s rotational inertia about the pivot pointpivot point

If θ is small If θ is small θ = sin θ θ = sin θ I

Lmg SHMSHM α ~ - (cte)·θα ~ - (cte)·θ

In figure above, as the pendulum moves to right its acceleration to the left In figure above, as the pendulum moves to right its acceleration to the left increases until it stops and starts moving to the left increases until it stops and starts moving to the left The movement of a simple The movement of a simple pendulum swinging through only small angles is approximately SHM.pendulum swinging through only small angles is approximately SHM.

Physical pendulum:

Simple pendulum angular acceleration: Simple pendulum angular acceleration: α α = = - ω- ω22 θ = - (Lmg/I)· θ θ = - (Lmg/I)· θ

TI

mgL 2

g

L

mgL

IT 22

I =m LI =m L2 2 rotational inertia of pendulum = simple pendulum rotational inertia of pendulum = simple pendulum particle (bob) at particle (bob) at radius L from pivot.radius L from pivot.

Difference with respect to simple pendulumDifference with respect to simple pendulum the restoring force component the restoring force component FFg g sinθsinθ has a has a

moment arm of distance moment arm of distance ““hh”” about the pivot point about the pivot point rather than the string length rather than the string length LL..

Complicated distribution of mass Complicated distribution of mass about the pivotabout the pivot

Small oscillation amplitudeSmall oscillation amplitude

mgh

IT 2

““II” depends on shape of physical pendulum ” depends on shape of physical pendulum about Oabout O

A physical pendulum will not swing if it pivots at its COM A physical pendulum will not swing if it pivots at its COM h=0 h=0 TT infinity infinity such pendulum will never complete one swing. such pendulum will never complete one swing.

Physical pendulum oscillating about O with period T can be represented Physical pendulum oscillating about O with period T can be represented by simple pendulum of length Lby simple pendulum of length L00 and same period. and same period.

Center of oscillation: Center of oscillation: point along the physical pendulum at distance Lpoint along the physical pendulum at distance L00

from O.from O.

A physical pendulum can be used to measure “g”.A physical pendulum can be used to measure “g”.

2

22

2

2

2

2

3

8

)2/(3

4)2(2

T

Lg

LM

ML

TMh

I

Tg

Mgh

IT

Example : pendulum = rod of length Example : pendulum = rod of length LL suspended suspended from one end.from one end.

h=L/2 (COM)

I = I (COM)+M hI = I (COM)+M h2 2 = ML= ML22 /12 + M (L/2) /12 + M (L/2)2 2 = ML= ML22/3/3

I (COM)=MLI (COM)=ML2 2 /12/12

VI. Simple Harmonic Motion and Uniform Circular MotionVI. Simple Harmonic Motion and Uniform Circular Motion

Simple harmonic motionSimple harmonic motion is the is the projection of uniform circular motionprojection of uniform circular motion on a on a diameter of the circle in which the later motion occurs.diameter of the circle in which the later motion occurs.

Particle P’Particle P’ uniform circular motion uniform circular motion φφ’= ’= ωωt + t + φφ

Particle PParticle P projection of projection of particle particle P’P’ onto x-axis onto x-axis x(t)= xx(t)= xmm cos ( cos (ωωt + t + φφ))

v = v = ωω·R ·R v = v = ωω·x·xmm

)sin()( txtv m

Projection of velocity of referenceProjection of velocity of referenceparticle = velocity SHMparticle = velocity SHM

Projection of radial acceleration of reference particle = Projection of radial acceleration of reference particle = acceleration SHMacceleration SHM

)cos()( 2 txta m

aarr = = ωω22·R ·R a = a = ωω22·x·xmm

VII. Damped Simple Harmonic MotionVII. Damped Simple Harmonic Motion

Damped motion Damped motion : : when the motion of an oscillator is reduced by an when the motion of an oscillator is reduced by an external force.external force.

ExampleExample: : Block oscillates vertically on a spring. From block, a rod Block oscillates vertically on a spring. From block, a rod extends to a vane submerged in liquid.extends to a vane submerged in liquid.

Liquid exerts a Liquid exerts a damping forcedamping force opposed to the opposed to the motionmotionbvFd

b = damping constant (kg/s)b = damping constant (kg/s)

Assumption:Assumption:

Gravitational force on block is negligible compared to the damping force Gravitational force on block is negligible compared to the damping force and the force on the block from the spring.and the force on the block from the spring.

02

2

kxdt

dxb

dt

xdmmakxbv

Solution:Solution: )'cos()( 2/ textx mbtm

Frequency of damped Frequency of damped oscillator:oscillator: 2

2

4'

m

b

m

k

m

k'If b=0 If b=0

(no damping)(no damping)

Mechanical energy Mechanical energy

Displacement of Displacement of damped oscillatordamped oscillator

mbtmexAmplitude 2/: Amplitude decreases exponentially with Amplitude decreases exponentially with

timetime

2

2

1)( mkxtE Damped Damped

oscillator:oscillator:mbt

mekxtE /2

2

1)( Un-damped Un-damped

oscillator:oscillator:

VIII. Forced oscillations and resonanceVIII. Forced oscillations and resonanceExamples:Examples:

1) Person swinging in a swing without anyone pushing 1) Person swinging in a swing without anyone pushing Free oscillationFree oscillation

2) Someone pushes the swing periodically 2) Someone pushes the swing periodically Forced or driven oscillationsForced or driven oscillations

Driven Driven oscillationsoscillations

Two frequencies:Two frequencies:

(1)(1) Natural angular frequencyNatural angular frequency ωω of the system of the system when system oscillates freely after when system oscillates freely after

a sudden disturbance.a sudden disturbance.

(2)(2) External frequencyExternal frequency ωωdd of the system of the system

angular frequency of the external driving forceangular frequency of the external driving forcecausing the driven oscillations.causing the driven oscillations.

If “rigid support” moves up and down If “rigid support” moves up and down forced forced simple harmonic oscillatorsimple harmonic oscillator

)cos()( txtx dm

XXmm depends on depends on ωωdd and and ωω

VelocityVelocity of oscillations greatest when: of oscillations greatest when: ωωdd = = ωω

Resonance Resonance Displacement Displacement amplitudeamplitude greatest greatest

 An oscillator consists of a block of mass 0.500 kg connected to a spring. When set into oscillation with amplitude 35.0 cm, the oscillator repeats its motion every 0.500 s. Find the (a) period, (b) frequency, (c) angular frequency, (d) spring constant, (e) maximum speed, and (f) magnitude of the maximum force on the block from the spring.

Solution• (a) The motion repeats every 0.500 s so the period must

be T = 0.500 s.• (b) The frequency is the reciprocal of the period: • f = 1/T = 1/(0.500 s) = 2.00 Hz.• (c) The angular frequency ω is ω = 2πf = 2π(2.00 Hz) =

12.6 rad/s.• (d) The angular frequency is related to the spring

constant k and the mass m by . We solve for k:• k = mω2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m.• (e) Let xm be the amplitude. The maximum speed is • vm = ωxm = (12.6 rad/s)(0.350 m) = 4.40 m/s.• (f) The maximum force is exerted when the displacement

is a maximum and its magnitude is given by Fm = kxm = (79.0 N/m)(0.350 m) = 27.6 N.

  In the figure below, two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg. The block is set oscillating on the frictionless floor. What is the frequency of oscillation?

Solution

• When displaced from equilibrium, the net force exerted by the springs is –2kx acting in a direction so as to return the block to its equilibrium position (x = 0). Since the acceleration a = d2x/dt2, Newton’s second law yields

• Substituting x = xm cos(ωt + φ) and simplifying, we find

= 2 . 2

2md x

dtkx

=2

2 k

m1 2 1 2(7580 N/m)

= = 39.6 Hz.2 2 2 0.245 kg

kf

m

A block weighing 14.0 N, which can slide without friction on an incline at angle θ = 40.0°, is connected to the top of the incline by a massless spring of unstretched length 0.450 m and spring constant 120 N/m. (a) How far from the top of the incline is the block's equilibrium point? (b) If the block is pulled slightly down the incline and

released, what is the period of the resulting oscillations?

Solution• (a) the equilibrium position when the block is gently

lowered until forces balance. If the amount the spring is stretched is x, then we examine force-components along the incline surface and find

• at equilibrium, The distance from the top of the incline is therefore (0.450 + 0.075) m = 0.525 m.

• (b) Just as with a vertical spring, the effect of gravity (or one of its components) is simply to shift the equilibrium position; it does not change the characteristics (such as the period) of simple harmonic motion.

14.0sin 40.0sin 0.0750 m

120kx mg x

14.0 9.80= 2 = 0.686 s.

120T