PH 221-2A Fall 2014

Oscillations

Lectures 22-23

Chapter 15 (Halliday/Resnick/Walker, Fundamentals of Physics 9th edition)

1

Chapter 15Oscillations

In this chapter we will cover the following topics:

Displacement, velocity and acceleration of a simple harmonic oscillator

Energy of a simple harmonic oscillator

Examples of simple harmonic oscillators: spring-mass system, simple pendulum, physical pendulum, torsion pendulum

Damped harmonic oscillator

Forced oscillations/Resonance2

OscillationsPeriodic motion – the motion of a particle or a system of particles is periodic, or cyclic, if it repeats again and again at regular intervals of time.

Example:• The orbital motion of a planet• The uniform rotational motion of a phonograph turntable• Back and forth motion of a piston in an automobile engine• Vibrations of a guitar string

Oscillation – back and forth or swinging periodic motion is called an oscillation

Simple Harmonic Motion

• Simple harmonic motion is a special kind of one dimensional periodic motion• The particle moves back and forth along a straight line, repeating the same motion again and again

Simple harmonic motion – the particles position can be expressed as a cosine or a sine function of time.

Cosines and sines are called harmonic functions => we call motion of the particle harmonic.

3

When an object attached to a horizontal spring is moved from its equilibrium position and released, the restoring force F = -kx leads to simple harmonic motion

Moving strip of paper at a steady rate

Record position of the vibrating object vs. time

pen

A = amplitude – the maximumdisplacement from equilibrium

x = 0 is the equilibrium position of the object.

Position as a function of time has the shape of trigonometric sine or cosine function

4

In fig.a we show snapshots of a sSimple Harmon

impleoscillat

ic

ing

Motio

syst

n (

em

SHM)

.

The motion is periodic i.e. it repeats in time. The time needed to completeone repetition is known as the The number of

repetitions per u

(symbol , units: s ).

nit time is called the f

perio

n

d

reque

T

cy

The displacement of the particle is given by the equation: Fig.b is a plot of ( ) versu amplitudes . The quantity is called the of the

(symbol

mot

, unit hertz, Hz) 1

( ) c

s

i

o

m

m

fT

x t x tx t

f

t x

o

n. It gives the maximum possible displacement of the oscillating objectThe quantity is called the of the oscillator. It is given by angular frequencythe equation:

( ) cosmx t x t

22 fT 5

( ) cosmx t x t The quantity is called the phase angle of the oscillator.The value of is determined from the displacement (0) and the velocity (0) at 0. In fig.a ( ) is plotted versus for 0. ( )

xv t x t

t x t x

Velocity of SHM

velocity

cos

( )( ) cos sin

The quantity is called the It expresses the maximum possible value of ( ) In fig.b the

am

velocity ( ) is plo

pl

tt

itude

m

m

m m

m

v

t

dx t dv t x t x tdt dt

xv t

v t

ed versus for 0. ( ) sinm

tv t x t

2 2

2

( )( ) sin cos

The quantity is called the

Acceleration of SHM:

acceleration ampli .It expresses the maximumpossible value of a( ). In fig.c the accele

tr

dat

u e am

m m

m

dv t da t x t x t xdt dt

xt

2

ion a( ) is plotted versus for 0. ( ) cosm

t ta t x t

6

2

2 2

We saw that the acceleration of an object undergoing SHM is:

If we apply Newton's second law we get:

The Force Law for Simple Harmonic Motion

Simple harmonic motion occurs whe

a x

F ma m x m x

The force can be writn the force acting on a

ten as: where is a constant. If we compare the two expressions for F we have:

n object is proportionalto the disaplacement but opposite in sign. F

CCx

2

d

anm C

Consider the motion of a mass attached to a springof srping constant than moves on a frictionless horizontal floor as shown in the figure.

mk

The net force on is given by Hooke's law: . If we compare this equation with the expression we identify the constant C with the sping constant k.We can then calculate the angular fre

F m F kxF Cx

quancy and the period .

and 2 2

T

C k m mTm m C k

km

2 mTk

2 mTC

Problem : When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12.0 Hz. When another object of mass m2 is hung on the spring along with m1, the frequency of the motion is 4.00 Hz. Find the ratio m2/m1 of the masses.

8

Three springs with force constants k1 = 10.0 N/m, k2 = 12.5 N/m, and k3 = 15.0 N/m are connected in parallel to a mass of 0.500 kg. The mass is then pulled to the right and released. Find the period of the motion.

9

A mass of 0.300 kg is placed on a vertical spring and the spring stretches by 10.0 cm. It is then pulled down an additional 5.00 cm and then released. Find:

a) K; b) ω; c) frequency; d) T; e) max velocity; f) amax;g) Fmax (max restoring force); h) V for x = 2.00 cm;i) The equations for displacement, velocity and acceleration at any time

10

11

12

The mechanical energy of a SHM is the sum of its potenEnergy in S

tial and kiimple Harmonic Motion

netic energies an

d

. E

U K

2 2 2

2 2 2 2 2 2

2 2 2 2

1 1Potential energy cos2 2

1 1 1Kinetic energy sin sin2 2 2

1 1Mechanical energy cos sin2 2

In the figure we plot the potential energy

m

m m

m m

U kx kx t

kK mv m x t m x tm

E U K kx t t kx

U

(green line), the kinetic energy (red line) and the mechanical energy (black line) versus time . While and

vary with time, the energy is a constant. The energy of the oscillating objectt

KE t U

K Eransfers back and forth between potential and kinetic energy, while the sum of

the two remains constant 13

Problem 35. A block of mass M =5.4 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a string of constant k=6000 N/m. A bullet of mass m=9.5 g and velocity

of magnitv yde 630 m/s strikes and is embedded in the block. Assuming the compression of the spring is negligible untill the bullet is embedded, determine (a) the speed of the blockimmidiately after the collison, and (b) the amplitude of the resulting simple harmonic motion

The problem consists of two distinct parts: the completely inelastic collision (which isassumed to occur instantaneously, the bullet embedding itself in the block before theblock moves through significant distance) followed by simple harmonic motion (of massm + M attached to a spring of spring constant k). (a) Momentum conservation readily yields v´ = mv/(m + M). With m = 9.5 g, M = 5.4 kg and v = 630 m/s, we obtain ' 1.1 m/s.v (b) Since v´ occurs at the equilibrium position, then v´ = vm for the simple harmonic motion. The relation vm = xm can be used to solve for xm, or we can pursue the alternate(though related) approach of energy conservation. Here we choose the latter:

2 22 2 2

21 1 1 1'2 2 2 2m m

m vm M v kx m M kxm M

which simplifies to

32

3

(9.5 10 kg)(630 m/s) 3.3 10 m.(6000 N/m)(9.5 10 kg 5.4kg)

mmvx

k m M

14

In the figure we show another type of oscillating systemIt consists of a disc of rotational

An Angular

inertia

Simple Harmonic Oscillator; Tors

suspended from a wire that tw

ion Pendul

ists as t

um

m roI

ates by an angle . The wire exertson the disc a This is the angular form of Hooke's law. The constant

is called th

restoring torque

torsion constane of

the wire t

.

If we compare the expression for the torque with the force equation we realize that we identify the constant C with the torsion constant .

We can thus readily determine the angular frequenF Cx

cy and the period of the

oscillation.

We note that I is the rotational inertia of the disc about an axis that coincides wit

2 2

h the wire. The angle is given b

C I ITI I C

T

y the equation:

( ) cosmt t 15

A simple pendulum consists of a particle of mass suspended by a string of length from a pivot point. If the mass is disturbed from its equilibrium position

Th

th

e Simple P

e net forc

endu

e ac

lu

n

m

ti g

mL

on it is such that the system exectutes simple harmonic motion.There are two forces acting on : The gravitational force and the tension from the string. The net torque of these forces is:

g

m

r F L sin Here is the angle that the thread

makes with the vertical. If 1 (say less than 5 ) then we canmake the following approximation: where is expressed in radians. Wit

sin h thi

mg

s approximation the torque is:

If we compare the expression for Lmg

with the force equation we realize that we identify the constant C with the term . We can thus readily determine the angular frequency

and the period of the oscillation. ; 2

F

C mgL TI

CxLmg

TI

2I IC mgL

In the small angle approximation we assumed that << 1 and used the approximation: sin We are now going to decide what is a “small” angle i.e. up to what angle is the approximation reasonably accurate?

(degrees) (radians) sin5 0.087 0.087

10 0.174 0.174

15 0.262 0.259 (1% off)

20 0.349 0.342 (2% off)

Conclusion: If we keep < 10 ° we make less that 1 % error

17

2

2

Physical Pendul

The rotational inertia about the pivot point is equal to

Thus 2 2

A physical pendulum is an extened rigid body that is suspendedfrom a fixed point O and oscill

um

ates

I mL

I mLTmgL mgL

under the influence of gravityThe net torque sin Here is the distance between point O and the center of mass C of the suspended body.If we make the small angle approximation 1, we have

mgh h

:

. If we compare the torques with the force equation we realize that we identify the constant C with the term . We can thus readily determine period of the oscillati

o

2

n.

2

mghF Cx

hmgT

ITC

Here is the rotational

inertia about an axis through O.

Imgh

I

2 LTg

2 ITmgh

18

Example 1: The pendulum can be used as a very simple device to measure the acceleration of gravity at a particular location.

• measure the length “l” of the pendulum and then set the pendulum into motion• measure “T” by a clock•

2 2 2 2

2 2

2

2

1 1 1( )12 2 3

For a uniform rod o

1/ 3 22 2 2 2 23 ( / 2) 3

83

f length L

comI I mh mL m L mL

I I mL L LTC mgh mgh g L g

LgT

19

1. From the piece of a rope and a bob we make a simple pendulum2. We set pendulum into motion3. We measure period “T” by a clock4. We calculate the length of the pendulum (rope)

5. With a help of the rope of the known length we measure the dimensions of the room a x b x h and its volume V = a x b x h

Example : The Length of a Pendulum. A student is in an empty room. He has a piece of a rope, a small bob, and a clock. Find the volume of the room.

20

Pr oblem 44. A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50 cm mark. The period of oscillation is 2.5 s. Find d?

We use Eq. 15-29 and the parallel-axis theorem I = Icm + mh2 where h = d, the unknown. For a meter stick of mass m, the rotational inertia about its center of mass is Icm = mL2/12 where L = 1.0 m. Thus, for T = 2.5 s, we obtain

T mL mdmgd

Lgd

dg

2 12 212

2 2 2

/ .

Squaring both sides and solving for d leads to the quadratic formula:

2 42 2/2 /2 /3= .

2g T g T L

d

Choosing the plus sign leads to an impossible value for d (d = 1.5 L). If we choose the minus sign, we obtain a physically meaningful result: d = 0.056 m.

2 ITmgd

21

Consider an object moving on a circular path of radius with a uniform speed . If we project th

Simple

e posit

Harmonic Motion and Unifor

ion of themoving particle

m Circular

at point

Moti

P on the

on

mxv

x-axis we get point P.

The coordinate of P is: ( ) cos . While point P executes unifrom circular motion its projection P moves along the x-axis with simple harmonic motion.The speed v of point

mx t x t

P is equal to . The direction of the velocity vector is along the tangentto the circular path. If we project the velocity on the x-axis we get: ( ) sin The acceleration points al

m

m

x

vv t x t

a

2

ong the center O. If we project along the x-axis we get: a( ) cos

Conclusion: Whether we look at the displacement ,the velocity, or the acceleration , the projection of uniform circular m

ma t x t

otion on the x-axis is SHM 22

When the amplitude of an oscillating object is reduced due to the presence of an external force the motion

Damped Simple Harmonic Motion

dampis said to be

An example is given in the figure. A med. ass m attached to a spring of spring constant k oscillates vertically.The oscillating mass is attached to a vane submerged in a

liquid. The liquid exerts a whose magnitu

damping force de is given by t

dF

he equation: dF bv

The negative sign indicates that opposes the motion of the oscillating mass.The parmeter is called the . The net force on m is:

damping From Newton's second

consta la nt

w we

d

net

Fb

F kx bv

2

2

2

2

have:

We substitute with and a with and get

the following differential equation: 0 d x dxm b kxd

dx d xkx bv ma vdt d

t t

t

d

23

2

2

Newton's second law for the damped harmonic oscillator:

0 The solution has the form:d x dxm b kxdt dt

/ 2

In the picture above we plot ( ) versus . We can regard the above solutionas a cosine function with a time-dependent amplitude . The angularfrequency of the damped harmonic oscillator

bt mm

x t tx e

2

22

is given by the equation:

1 For an undamped harmonic oscillator the energy 2

If the oscillator is damped its energy is not constant but decreases with time. If the

k=m

damping is s

4 mE kxbm

/ 2

2 /

mall we can replace with By doing so we find that:1( ) The mechanical energy decreases exponentially with time 2

bt mm m

bt mm

x x e

E t kx e

/ 2 ( ) cosbt mmx t x e t

24

25

Problem 59. The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?

Since the energy is proportional to the amplitude squared (see Eq. 15-21), we find the fractional change (assumed small) is

= = 2 = 2 . 2

2 2

E E

EdEE

dxx

x dxx

dxx

m

m

m m

m

m

m

Thus, if we approximate the fractional change in xm as dxm/xm, then the above calculation shows that multiplying this by 2 should give the fractional energy change. Therefore, ifxm decreases by 3%, then E must decrease by 6.0 %.

212 mE U K kx

Moving support

If an oscillating system is disturbed and then allowedto oscillate freely the corresponding angular frequency is calle

Forced Oscillations and Resonance

nad the The satural f me systrequ em ce ancy l. n aso

be driven as shown in the figure by a moving support thatoscillates at an arbitrary angular frequency . Such a forced oscillator oscillates at the angular frequency of the driving force. The

d

d

displacement is given by:

The oscillation amplitude varieswith the driving frequency as shown in the lower figure.The amplitude is approximatetly greatest when

( ) co

This condition

s m

d

mx t x xt

is called All mechanical structureshave one or more natural frequencies and if a structure issubjected to a strong external driving force whose frequencymatches one of the natura

resona

l fre

nce.

quencies, the resulting oscillationsmay damage the structure 26

Problem 63. A 100 kg car carrying four 82 kg people travels over a "washboard" dirt roadwith corrugations 4.0 m apart. The car bounces with maximum amplitude when its speed is 16 km/h. When the car stops, and the people get out, how much does the car body rise on its suspension?

With M = 1000 kg and m = 82 kg, we adapt Eq. 15-12 to this situation by writing

24

kT M m

.

If d = 4.0 m is the distance traveled (at constant car speed v) between impulses, then we may write T = d/v, in which case the above equation may be solved for the spring constant:

22 2= 4 .

4v k vk M m

d M m d

Before the people got out, the equilibrium compression is xi = (M + 4m)g/k, and afterward it is xf = Mg/k. Therefore, with v = 16000/3600 = 4.44 m/s, we find the rise ofthe car body on its suspension is

= 4 = 4+ 4 2

= 0.050 . 2

x x mgk

mgM m

dvi f F

HIK m

km

27

28

We wish to find the effective spring constant for the combination of springs. We do this by finding the magnitude F of the force exerted on the mass when the total elongation ofthe springs is x. Then keff = F/x. Suppose the left-hand spring is elongated by x and the right-hand spring is elongated by xr. The left-hand spring exerts a force ofmagnitude k x on the right-hand spring and the right-hand spring exerts a force ofmagnitude kxr on the left-hand spring. By Newton’s third law these must be equal, so x xr . The two elongations must be the same and the total elongation is twice theelongation of either spring: x x 2 . The left-hand spring exerts a force on the block and its magnitude is F k x . Thus k k x x kreff / /2 2 . The block behaves as if it were subject to the force of a single spring, with spring constant k/2. To find the frequency of its motion replace keff in f k m 1 2/ /a f eff with k/2 to obtain

= 12 2

f km

.

With m = 0.245 kg and k = 6430 N/m, the frequency is f = 18.2 Hz.

Problem 26. Two springs are joined in series and connected to a block of mass 0.245 kgthat is set oscillating over a frictionless floor. The springs each have spring constant k=6430 N/m. What is the frequency of oscillations?