PH 221-3A Fall 2010

Oscillations

Lectures 23-24

Chapter 15 (Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)

1

Chapter 15OscillationsOscillations

In this chapter we will cover the following topics:

Displacement, velocity and acceleration of a simple harmonic oscillatorDisplacement, 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 motionp p p• 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 p p p ptime.

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 objectposition of the object.

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

4

Simple Harmonic Motion (SHM)

( ) cosmx t x t

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 (symbol , units: s ). period T

1repetitions per unit time is called the f nreque

cy

The displacement of the particle is given by the equation:

(symbol , unit hertz, Hz) 1

( ) c som

fT

x t x t

f

Fig.b is a plot of ( ) versu amplitudes . The quantity is called the of themot

i

mx t t xon. It gives the maximum possible displacement of the oscillating object

The quantity is called the of the oscillator It is given byangular frequency

The quantity is called the of the oscillator. It is given by angular frequencythe equation:

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) x

and the velocity (0) at 0. In fig.a ( ) is plotted versus for 0. ( )

v t x tt x t x

Velocity of SHMcosm t

Velocity of SHM( )( ) cos sinm m

dx t dv t x t x tdt dt

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

am

velocity ( ) is plo

pl

tt

itude m mvxv t

v t

ed versus for 0t In fig.b the velocity ( ) is plottv t ed versus for 0. ( ) sinm

tv t x t

2 2( )dv t d 2 2

2

( )( ) sin cos

The quantity is called the

Acceleration of SHM:

acceleration ampli .It expresses the maximumt du e am

m m

m

dv t da t x t x t xdt dt

x

possible value of a( ). In fig.c the acceleratt2

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: The Force Law for Simple Harmonic Motion

a x

2 2If we apply Newton's second law we get:

Simple harmonic motion occurs whe

F ma m x m x

The force can be writn the force acting on a

ten as:n object is proportional

to the disaplacement but opposite in sign F Cx The force can be written as: where is a constant. If we compare the two expressions for F we have:to the disaplacement but opposite in sign. F

CCx

2

d

anm C 2 mT

Consider the motion of a mass attached to a springf i t t th f i ti l

mk

2TC

of srping constant than moves on a frictionless horizontal floor as shown in the figure.

k

The net force on is given by Hooke's law: If we compare this equationF m F kxThe 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 .T

and 2 2C k m mTm m C k

km

2 mTk

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;) F ( i f ) h) V f 2 00g) Fmax (max restoring force); h) V for x = 2.00 cm;

i) The equations for displacement, velocity and acceleration at any time

10

11

Problem 24

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 21 1P i l U k k

2 2 2

2 2 2 2 2 2

Potential energy cos2 2

1 1 1Kinetic energy sin sin2 2 2

m

m m

U kx kx t

kK mv m x t m x tm

2 2 2 2

2 2 21 1Mechanical energy cos sin2 2

I h fi l h i l

m m

m

E U K kx t t kx

U

( li ) h ki i KIn the figure we plot the potential energy 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 object

KE t U

K Ey , gy gy g jtransfers 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 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 isp p p y (assumed 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 ( ) q p , m pmotion. 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 2

2 2 21 1 1 1m v

2 2 22

1 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 systemAn Angular Simple Harmonic Oscillator; Torsion PendulumIn the figure we show another type of oscillating systemIt consists of a disc of rotational inertia suspended from a wire that twists as tm ro

Iates by an angle . The wire exerts

on the disc a This is the angular form of Hooke's law. The constant

is called th

restoring torque

torsion constane of

the wiret

is called th torsion constane of the wire t .

If we compare the expression for the torque with the force equation

p p q q 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 theT

oscillation.

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

2 2

h

C I ITI I C

We note that I is the rotational inertia of the disc about an axis that coincides with the wire. The angle is given b 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

The Simple Pendulumm

Lby a string of length from a pivot point. If the mass is disturbed from its equilibrium position the net force ac nti g

L

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:

m

F L i H i th l th t th th d gr 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 sin

mg

g pp

expressed in radians. With this

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

F CxLmg

and the period of the oscillation. ; 2C mgL TI

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” pp g gangle 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)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

The rotational inertia about the pivot point is equal to I mL

I mL

Physical Pendul

Thus 2 2

um

I mLTmgL mgL

A physical pendulum is an extened rigid body that is suspendedfrom a fixed point O and oscillates under the influence of gravityThe net torque sin Here is the distance between mgh h

2 LTg

point O and the center of mass C of the suspended body.If we make the small angle approximation 1, we have:

If we compare the torques with themgh . If we compare the torques with the force equation we realize that we identify the constant C with the term . We can thus readily

mghF Cx

hmg

determine period of the oscillatio

2 2

n.

I ITC

T

Here is the rotational h

Img

Iinertia about an axis through O.

g 2 IT

mgh

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•

For a uniform rod of length L

2 2 2 21 1 1( )12 2 3

g

comI I mh mL m L mL

2 2

2

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

I I mL L LTC mgh mgh g L g

2

2

83

LgT

19

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

1. From the piece of 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)

i h h l f h f h k l h h5. 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

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. Fi d d?Find d?

2 ITmgd

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 / 2 / 3

2.

2 2 4 2

dg T d T L a f a f

2 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.

21

Consider an object moving on a circular path of radius Simple Harmonic Motion and Uniform Circular Motion

mxwith a uniform speed . If we project the position of themoving particle at point P on the

v

x-axis we get point P.

The coordinate of P is: ( ) cos . mx t x t ( )While point P executes unifrom circular motion its projection P moves along the x-axis with simple h i ti

m

harmonic motion.The speed v of point P is equal to . The direction of the velocity vector is along the tangent

mx

to the circular path. If we project the velocity on the x-axis we get: ( ) sin Th l ti i t l

m

vv t x t

th t O If j tThe acceleration points ala

2

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

Conclusion: Whether we look at the displacement ,the ma t x t

velocity, or the acceleration , the projection of uniform circular motion on the x-axis is SHM 22

When the amplitude of an oscillating object is reduced dueDamped Simple Harmonic Motion

to the presence of an external force the motiondamp

is said to be An example is given in the figure. A med. ass m

attached to a spring of spring constant k oscillates verticallyattached 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 damping force dF

magnitude is given by the equation: dF bv

The negative sign indicates that opposes the motion of the oscillating mass.dF

g g pp gThe parmeter is called the . The net force on m is:

damping From Newton's second

consta la nt

w we

d

net

bF kx bv have:

2

2

2

We substitute with and a with and get

d x dx

dx d xkx bv ma vdt dt

2 the following differential equation: 0d x dxm b kxdt td

23

Newton's second law for the damped harmonic oscillator:2

2 0 The solution has the form:d x dxm b kxdt dt

/ 2( ) cosbt mx t x e t

/ 2

In the picture above we plot ( ) versus . We can regard the above solutionas a cosine function with a time-dependent amplitude The angularbt m

x t tx e

( ) cosmx t x e t

as a cosine function with a time dependent amplitude . The angularfrequency of the damped harmonic oscillator

mx e

22

is given by the equation:

1For an undamped harmonic oscillator the energyk= E kxb 2 For an undamped harmonic oscillator the energy 2

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

m

damping is s

4 mE kxm

/ 2mall we can replace with By doing so we find that:bt mm mx x ep g

2 /

p y g1( ) The mechanical energy decreases exponentially with time 2

m m

bt mmE t kx e 24

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?

21

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

212 mE U K kx

fractional change (assumed small) is

= = 2 = 2 . 2

2 2

E E

EdEE

dxx

x dxx

dxx

m

m

m 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, ifx decreases by 3% then E must decrease by 6 0 %xm decreases by 3%, then E must decrease by 6.0 %.

25

Moving support

If an oscillating system is disturbed and then allowedForced Oscillations and Resonance

to oscillate freely the corresponding angular frequency is calle nad the The satural f me systrequ em ce ancy l. n a

so

be driven as shown in the figure by a moving support thatbe 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

d

d

of the driving force. The

displacement is given by:

The oscillation amplitude variesi h h d i i f h i h l fi( ) cos mmx t x xt

with the driving frequency as shown in the lower figure.The amplitude is approximatetly greatest when This condition

d is called All mechanical structuresresonance.This condition is called All mechanical structures

have one or more natural frequencies and if a structure issubjected to a strong external driving force whose frequency

resonance.

matches one of the natural frequencies, 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 rise16 km/h. When the car stops, and the people get out, how much does the car body rise on its suspension?

With M 1000 k d 82 k d t E 15 12 t thi it ti b iti

km

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 andBefore 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

di f F

HIK m

+ 4 2k M m vf H K27

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

We wish to find the effective spring constant for the combination of springs shown in the

k=6430 N/m. What is the frequency of oscillations?

We wish to find the effective spring constant for the combination of springs shown in the figure. We do this by finding the magnitude F of the force exerted on the mass when thetotal elongation of the 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 of magnitude k x on the right-hand spring and the right-hand spring exerts a force of magnitude kxr on the left-hand spring. By Newton’s third law these must beequal, so x xr . The two elongations must be the same and the total elongation istwice the elongation of either spring: x x 2 . The left-hand spring exerts a force ontwice the elongation of either spring: x x2 . The left hand spring exerts a force onthe 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

.

28

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