OSCillATIONS AND WAVES

Oscillations

14-1 Simple Harmonic Motion

14-2 Energy in Simple Harmonic Motion

14-3 Some Oscillating Systems

14-4 Damped Oscillations

14-5 Driven Oscillations and Resonance

e discuss oscillatory motion in this chapter. The kinematics of motionwith constant acceleration is presented in Chapters 2 and 3. In thischapter, the kinematics and dynamics of motion with accelerationthat is proportional to displacement from equilibrium is presented. Theword "oscillate" means to swing back and forth. Oscillation occurswhen a system is disturbed from a position of stable equilibrium. Many

familiar examples exist: surfers bob up and down waiting for the right wave, clockpendulums swing back and forth, and the strings and reeds of musical instrumentsvibrate.

Other, less familiar examples are the oscillations of air molecules in a soundwave and the oscillations of electric currents in radios, television sets, and metaldetectors. In addition, many other devices rely on oscillatory motion to function.

In this chapter, we deal mostly with the most fundamental type of oscillatorymotion-simple harmonic motion. We also consider both damped anddriven oscillations.

MONSTER TRUCKS CAN POWER OVERJUST ABOUT ANYTHING, BUT WHATKEEPSTHESE GIANTTRUCKS FROMTHROWING THEIR DRIVERS RIGHT OUTOFTHEIR SEATS? MONSTER TRUCKSHAVE MONSTER-SIZE SHOCKABSORBERS. THESE GIANT SHOCKABSORBERS HELP DAMPENTHEOSCillATION OFTHE VEHICLE,PROVIDING A SMOOTHER RIDE ASTHEOPERATOR DRIVES OVERTOUGHTERRAIN OR EVEN OTHERTRUCKS.(Jeff GreenbergIPhotoedit.)

How does a mechanic installing

monster truck shock absorbers

determine which size shock

absorber to use? (See Example 14-13.)

14-1 SIMPLE HARMONIC MOTIONA common, very important, and very basic kind of oscillatory motion is simple har-monic motion such as the motion of a solid object attached to a spring (Figure 14-1).In equilibrium, the spring exerts no force on the object. When the object is displacedan amount x from its equilibrium position, the spring exerts a force -kx, as given by IHooke's law:* ~

where k is the force constant of the spring, a measure of the spring's stiffness. Theminus sign indicates that the force is a restoring force; that is, it is oppositeto the direction of the displacement from the equilibrium position. CombiningEquation 14-1 with Newton's second law (F

y= ma), we have

-kx = max

ka = --x

x m ( or

The acceleration is proportional to the displacement and the minus sign indicatesthat the acceleration and the displacement are oppositely directed. This relation isthe defining characteristic of simple harmonic motion and can be used to identifysystems that will exhibit it:

In simple harmonic motion, the acceleration, and thus the net force, areboth proportional to, and oppositely directed from, the displacement fromthe equilibrium position.

The time it takes for a displaced object to execute a complete cycle of oscillatorymotion-from one extreme to the other extreme and back-is called the period T.The reciprocal of the period is the frequency j, which is the number of cycles perunit of time:

The unit of frequency is the cycle per second (cyjs), which is called a hertz (Hz).For example, if the time for one complete cycle of oscillation is 0.25 s, the frequencyis 4.0 Hz.

Figure 14-2 shows how we can experimentally obtain x versus t for a mass on aspring. The general equation for such a curve is

where A, w, and 8 are constants. The maximum displacement xmax from equilib-rium is called the amplitude A. The argument of the cosine function, wt + 8, is

,~~x----.I,

FIG U R E 1 4 - 1 An object and spring on africtionless surface. The displacement x,measured from the equilibrium position, ispositive if the spring is stretched and negativeif the spring is compressed.

FIG U R E 1 4·2 A marking pen is attachedto a mass on a spring, and the paper is pulledto the left. As the paper moves with constantspeed, the pen traces out the displacement xas a function of time t. (Here, we have chosen xto be positive when the spring is compressed.)

=:.......i'dthe phase of the motion, and the constant 8 is called the phase constant,_-:..d1 equals the phase at t = O. [Note that cos(wt + 8) = sin(wt + 8 + [7T/2]);

_ ~, whether the equation is expressed as a cosine function or a sine function sim-- _-- epends on the phase of the oscillation at t = 0.] If we have just one oscillating

:o:em,we can always choose t = 0 so that 8 = o. If we have two systems oscillat-_ ~ ,,-ith the same frequency but with different phases, we can choose 8 = 0 for one- :hem. The equations for the two systems are then

Xl = Al cos(wt)

x2 = A2 cos(wt + 8)

e phase difference 8 is 0 or an integer times 27T, then the systems are said to be'Jhase. If the phase difference 8 is 7T or an odd integer times 7T, then the systems

.:...c said to be 1800 out of phase.We can show that Equation 14-4 is a solution of Equation 14-2 by differentiating

:wice with respect to time. The first derivative of X gives the velocity vx:

dx-, = - = -wA sin(wt + 8)

x dt

Jifferentiating velocity with respect to time gives the acceleration:

dv d2xa = _x = - = -w2A cos(wt + 8) 14-6

x dt dt2

Substituting X for A cos(wt + 8) (see Equation 14-4) gives

Comparing ax = -w2x (Equation 14-7) with at = -(k/m)x (Equation 14-2), we seethat x = A cos(wt + 8) is a solution of d2x/dt2 = -(k/m)x (Equation 14-2) if

w= (k\j-;;;The amplitude A and the phase constant 8 can be determined from the initial po-sition Xo and the initial velocity vox of the system. Setting t = 0 in x = A cos(wt + 8)

·ves

Xo = A cos8

Similarly, setting t = 0 in Vx = dx/dt = - Aw sin(wt + 8) gives

By using these equations, we can determine A and 8 in terms of XO' vax' and w.The period T is the shortest time interval satisfying the relation

for all t. Substituting into this relation using x(t) = A cos(wt + 8) (Equation 14-4)gives

A cos(wt + 8) = A cos[w(t + T) + 8]= A cos(wt + 8 + wT)

The swaying of the Citicorp Building in NewYork City during high winds is reduced bythis tuned-mass damper mounted on anupper floor. It consists of a 400-ton slidingblock connected to the building by a spring.The force constant is chosen so that thenatural frequency of the spring-block systemis the same as the natural sway frequency ofthe building. Set into motion by winds, thebuilding and damper oscillate 1800 out ofphase with each other, thereby significantlyreducing the swaying. (Citibank.)

SeeMath Tutorial for more

information on

The cosine (and sine) function repeats in value when the phase increases by21T, so

wT = 21T or W = 21T (t)The constant w is called the angular frequency. It has units of radians per secondand dimensions of inverse time, the same as angular speed, which is also designatedby w. Substituting 21T/T for w in Equation 14-4 gives

We can see by inspection that each time t increases by T, the ratio tiT increases by1, the phase increases by 21T, and one cycle of the motion is completed.

The frequency is related to the angular frequency by

Because w = Vkj;z, the frequency and period of an object on a spring are relatedto the force constant k and the mass In by

The frequency increases with increasing k (spring stiffness) and decreases withincreasing mass. Equation 14-12 provides a way to measure the inertial mass of anastronaut in a "weightless" environment.

PRACTICE PROBLEM 14-1A o.So-kg object is attached to a spring that has a force constant k = 400 N/m. (a) Find thefrequency and period of motion of the object when it is displaced from equilibrium andthen released. (b) Repeat Part (a) except with a 1.6-kg object attached to the spring in placeof the o.So-kg object. Hint: Review Example 14-4 first.

PICTURE Choose the origin of the x axis at the equilibrium position.For a spring, choose the +x direction so that x is positive if the spring isextended.

SOLVE Do not use the kinematic equations for constant acceleration. Instead,use the equations developed for simple-harmonic motion.

CHECK Make sure your calculator is in the appropriate mode(degrees or radians) when evaluating trigonometric functions and theirarguments.

At just 1 rad/s off resonance, theamplitude drops by a factor of 20.

This is not surprising, because thewidth ~w of the resonance is only0.0957 rad/s.

Astronaut Alan 1. Bean measures his bodymass during the second Skylab mission bysitting in a seat attached to a spring andoscillating back and forth. The total massof the astronaut plus the seat is related tohis frequency of vibration by Equation 14-12.(NASA.)

___ Riding the Waves

_"u are sitting on a surfboard that is riding up and down on some swells. The board's--"rtical displacement y is given by

y = (1.2 m)cos(2.~ s t + ~)

.• Find the amplitude, angular frequency, phase constant, frequency, and period of the::-,otion. (b) Where is the surfboard at t = 1.0 s? (c) Find the velocity and acceleration as::-~ctions of time t. (d) Find the initial values of the position, velocity, and acceleration of the=fboard.

:>ICTURE We find the quantities asked for in (a) by comparing the equation of motion

y = (1.2 m) cos(-Ol t + ~)2. s 6

- ;th the standard equation for simple harmonic motion, Equation 14-4. The velocity and.=. celeration are found by differentiating y(t).

.•) 1. Compare this equation with y = A cos(wt + 0)(Equation 14-4) to get A, w, and 0:

l") Set t = 1.0 s to find the surfboard's position abovemean sea level:

) The velocity and acceleration are obtained from theposition by differentiation with respect to time:

( 1 'iT)Y = (1.2 m) cos 2.0 s t + Ii

A = 11.2 m 1 w =I0.50 rad/s I 0 =8w 0.50 rad/s I If = 27T= 27T = 0.0796 Hz = 0.080 Hz

T = .!. = 1 = 12.6 s = ~f 0.0796 Hz L::::.:J

y = (1.2 m) cos[ (0.50 rad/s)(1.0 s) + ~] = I 0.62 m I

dy dv = -d = -d [A cos(wt + 0)] = -wA sin(wt + 0)

y t t

= -(0.50 rad/s)(1.2 m) sin[(0.50 rad/s)t + ~ ]

= -(0.60 m/s) Sin[(0.50 rad/s)t + ~ ]

dv day = d: = dt [-wA sin(wt + 0)] = -w2 A cos(wt + 0)

= - (0.50 rad/s)2(1.2 m) cos[ (0.50 rad/s)t + ~ ]

= -(0.30 m/s2) cos[(0.50 rad/s)t + ~]

Yo = (1.2m)cos~ = 1.04 =~

VOy = -(0.60 m/s) sin ~ = 1-0.30 m/s IaOy = -(0.30m/s2)cos~ = 1-0.26m/s21

CH ECK We can check the plausibility of the Part (d) results using ay = -w2y (Equation 14-7)at t = 0, with Y = 1.04 m and w = 0.50 rad/s. Substituting into Equation 14-7 gives.l:'V = -w2yo = -(0.50 rad/s)2(1.04 m) = ~0.26 m/s2, the same as the third Part (d) result.

Figure 14-3 shows two identical masses attached to identi-cal springs and resting on a horizontal frictionless surface. Thespring attached to object 2 is stretched 10 cm and the spring at-tached to object 1 is stretched 5 cm. If they are released at thesame time, which object reaches the equilibrium position first?

According to Equation 14-12, the period depends only onk and m and not on the amplitude. Because k and m are thesame for both systems, the periods are the same. Thus, theobjects reach the equilibrium position at the same time. Thesecond object has twice as far to go to reach equilibrium, butit will also have twice the speed at any given instant. Figure14-4 shows a sketch of the position functions for the two ob-jects. This sketch illustrates an important general property ofsimple harmonic motion:

The frequency (and thus the period) of simple har-monic motion is independent of the amplitude.

The fact that the frequency in simple harmonic motion isindependent of the amplitude has important consequences inmany fields. In music, for example, it means that when a noteis struck on the piano, the pitch (which corresponds to thefrequency) does not depend on how loudly the note is played(which corresponds to the amplitude).t If changes in ampli-tude had a large effect on the frequency, then musical instru-ments would be unplayable.

Example 14-2FIG U R E 1 4 - 4 Plots of x versus t for the systems in Figure 14-3.Both reach their equilibrium positions at the same time.

An object oscillates with angular frequency w = 8.0 rad/s. At t = 0, the object is atx = 4.0 cm with an initial velocity vt = -25 cm/s. (a) Find the amplitude and phase constantfor the motion. (b) Write x as a function of time.

PICTURE The initial position and velocity give us two equations from which to determinethe amplitude A and the phase constant 8.

SOLVE

(a) 1. The initial position and velocity are related to theamplitude and phase constant. The position is given byEquation 14-4. The velocity is found by taking thederivative with respect to time:

2. At t = 0 the position and velocity are:

x = A cos(wt + 8) and

dxv = -d = -wA sin(wt + 8)

x t

-wAsin8A ~ = -wtan8

coso

VOxtan8 = -- so

wxo

(Vox) [ -25 cm/s ]8 = tan-1

- wxo = tan-1- (8.0 rad/s)(4.0 cm)

= 0.663 rad = I 0.66 rad I

t For many musical instruments, there is a slight dependence of frequency on amplitude. The vibration of an oboe reed,for example, is not exactly simple harmonic; thus its pitch depends slightly on how hard it is blown. This effect can becorrected for by a skilled musician.

~ The amplitude can be found using either the Xo or voxequation. Here we use xo:

A=l= 4.0cm =15.1cmlcos 8 cos 0.663 ---

x = I (5.1 cm)cos[(8.0 S-l)t + 0.66]1

- ECK To see if the Part-(b) result (x = (5.1 em) cos[(8.0 S-l)t + 0.66]) is plausible, we set t-~ to zero and see if x = 4.0 em. That is, x = (5.1 em) cos[(O) + 0.66] = 4.0 em. Thus, the~ "-- -(b) result is plausible.

x = A cos wt

Vx = -wA sin wt14-13a14-13b

ax = -w2A cos wt

-=-.: ese functions are plotted in Figure 14-5.

FIG U R E 1 4 - 5 Plots of x, Vx and ax as functions of time t for8 = O.At t = 0, the displacement is maximum, the velocity iszero, and the acceleration is negative and equal to -w2A. Thevelocity becomes negative as the object moves back toward itsequilibrium position. After one quarter-period (t = T/ 4), theobject is at equilibrium, x = 0, ax = 0, and the velocity has itsminimum value of -wA. At t = T/2, the displacement is - A, thevelocity is again zero, and the acceleration is +w2 A. Att = 3T/4, x = O,ax = 0, and Vx = +wA.

_ A Block on a Spring

.-'>. 2.00-kg block is attached to a spring as in Figure 14-1. The force constant of the spring isi: = 196 N/m. The block is held a distance 5.00 em from the equilibrium position and is re-::- ~ed at t = O. (a) Find the angular frequency w, the frequency f and the period T. (b) Write: ~ a function of time.

T T 3T4" 2" 4" T

OLVE

Cover the column to the right and try these on your own before looking at the answers.

teps

.<) 1. Calculate w from w = ~. w = 19.90 rad/s If = 11.58 Hz I T = I 0.635 s IA = 5.00 em 8 = 0.00

x = I (5.00 cm)cos[(9.90 S-l)t] I2. Use your result to find f and T.

3. Find A and 8 from the initial conditions.

HECK The block was released from rest, so we expect the velocity at t = 0 to be zero.-:-0 verify that our Part-(b) result is correct, we take the derivative of the expressionI = (5.00 cm)cos[(9.90 S-l)t] and evaluate it at t = O. That is, vJt) = dx/dt =-(4.95 cm/s)sin[(9.90 S-l)t]. Evaluating this aU = 0 gives v/O) = -(4.95 cm/s) sin(O) = 0, asi"<pected.

Example 14-4Consider an object on a spring whose position is given by x = (S.OOcm)cos(9.90 s-lt).(a) What is the maximum speed of the object? (b) When does this maximum speed first occurafter I = O? (c) What is the maximum of the acceleration of the object? (d) When does themaximum of the magnitude of the acceleration first occur after I = O?

PICTURE Because the object is released from rest, 15 = 0, and the position, velocity, andacceleration are given by Equations 14-13a, b, and c.

SOLVE

(a) 1. Equation 14-13a, with 15 = 0, gives the position. We getthe velocity by taking the derivative with respect to time:

x = A coswldx

so v = -d = - wA sin wtx I

v = wAlsin wtlso vrnax = wA = (9.90 rad/s)(S.OO cm)

= 149.5 cm/s I71"371"S71"

Isinwtl = 1 => wt = 2'2'2""

I = 2: = 2(9.9~ S-I) = I 0.lS9 s Idv

ax = _x = -w2A cos wtdt

(c) 1. We find the acceleration by taking the derivative of thevelocity, obtained in step 1 of Part (a):

arnax = w2A = (9.90 rad/s)2(S.00 cm) = 1490 cm/s2 "" !glI = ~ = __ 71"_ = I 0.317 s I

w 9.90s-1 ---,(d) The magnitude of the acceleration is maximum when

Icos wll = 1, which is when wt = 0,71",271",... :

CHECK We expect laxl to first be maximum after t = 0 when x reaches its first minimum,and we expect x to reach its first minimum one-half cycle after release. That is, we expect laxlto be maximum when I = !T, where T is the period. The period and the angular frequencyare related by w = 271"f = 271"/T (Equation 14-11). Substituting 271"/T for w in our Part (d)result gives t = 71"/(271"/T) = !T, as expected.

A relation exists between simple harmonic motion and circular motion with con-stant speed. Imagine a particle moving with constant speed v in a circle of radius A(Figure 14-6a). Its angular displacement relative to the +x direction is given by

FIG U R E 14 - 6 A particle moves in acircular path with constant speed. (a) Its xcomponent of position describes simpleharmonic motion, and (b) its x component ofvelocity describes the velocity of the simpleharmonic motion.

-"':-e8 is the angular displacement at time t = 0 and w = viA is the angular~ of the particle. The x component of the particle's position (Figure 14-6b) is

- nen a particle moves with constant speed in a circle, its projection onto a:iiameter of the circle moves with simple harmonic motion (see Figure 14-6).

---2 speed of a particle moving in a circle is rw, where r is the radius. For the--~cle in Figure 14-6b, r = A, so its speed is Aw. The projection of the velocity

=-.::or onto the x axis gives Vx = -v sin e. Substituting for v and e gives

_-ich is the same as Equation 14-5 for simple harmonic motion. The relation_~:-.,·eencircular motion and simple harmonic motion is nicely demonstrated by-.,. age of the bubble trail produced by a rotating boat propeller.

ENERGY IN SIMPLE HARMONIC MOTION-: en an object on a spring undergoes simple harmonic motion, the system's po-

-- .tial energy and kinetic energy vary with time. Their sum, the total mechanical_ ergy E = K + U, is constant. Consider an object a distance x from equilibrium,: --:2don by a restoring force -kx. The system's potential energy is

- is Equation 7-4. For simple harmonic motion, x = A cos(wt + 8). Substituting=\-es

-~ere m is the object's mass and v is its speed. For simple harmonic motion,-wA sin(wt + 8). Substituting gives

-=- e total mechanical energy E is the sum of the potential and kinetic energies:

E = U + K = !kN cos2(wt + 8) + !k A sin2(wt + 8)

= !kN[cos2(wt + 8) + sin2(wt + 8)]

Bubbles foaming off the edge of a rotatingpropeller that is moving through waterproduce a sinusoidal pattern. (Institute forMarine Dynamics.)

E = U + K = ~kA2

TOTAL MECHANICAL ENERGY IN SIMPLE HARMONIC MOTION

The total mechanical energy in simple harmonic motion is proportional tothe square of the amplitude.

For an object at its maximum displacement, the total energyis all potential energy. As the object moves toward its equi-librium position, the kinetic energy of the system increasesand its potential energy decreases. As the object movesthrough its equilibrium position, the kinetic energy of the ob-ject is maximum, the potential energy of the system is zero,and the total energy is kinetic.

As the object moves past the equilibrium point, its kineticenergy begins to decrease, and the potential energy of thesystem increases until the object again stops momentarily atits maximum displacement (now in the other direction). Atall times, the sum of the potential and kinetic energies is con-stant. Figure 14-7b and c show plots of U and K versus time.These curves have the same shape except that one is zerowhen the other is maximum. Their average values over oneor more cycles are equal, and because U + K = E, their aver-age values are given by

In Figure 14-8, the potential energy U is graphed as a functionof x. The total energy E is constant and is therefore plotted asa horizontal line. This line intersects the potential-energycurve at x = A and x = - A. At these two points, called theturning points, oscillating objects reverse direction and headback toward the equilibrium position. Because U $ E, themotion is restricted to - A $ X $ +A.

,,, ,, ,, ,, ,, ,- - - -:- - - - -: Etotal, ,, ,, ,

1 -: Kav = !Etotal, ,, ,, ,, ,, 0

- - -,- - - - -,- - - -, - - - - T - - - - r - - - -r - - - -,-- --I I I I II I I I II I I I II I I I I

,,,,,,,,- - - -,- - --,,,,

FIG U R E 1 4 - 8 The potential-energyfunction U = ~kx2 for an object of mechanicalmass m on a (massless) spring of force constantk. The horizontal blue line represents the totalmechanical energy E'o'ol for an amplitude of A.The kinetic energy K is represented by thevertical distance K = E'o'al - U. E,o'ol 2: U,SOthe motion is restricted to - A :$ X :$ +A.

,0,,,,---I ,,,,,,,

&ample 14-5.-. 3.0-kg object attached to a spring oscillates with an amplitude of 4.0 cm and a period of- 0 s. (a) What is the total energy? (b) What is the maximum speed of the object? (c) At what?Osition Xl is the speed equal to half its maximum value?

ICTURE (a) The total energy can be found from the amplitude and the force constant, and:::e force constant can be found from the mass and period. (b) The maximum speed occurs..;' en the kinetic energy equals the total energy. (c) We can relate the position to the speed=y using conservation of energy.

:;) 1. Write the total energy E in terms of the force constant kand amplitude A:

2. The force constant is related to the period and mass:

b) To find vrnax' set the kinetic energy equal to the totalenergy and solve for v:

c) 1. Conservation of energy relates the position X tothe speed v:

2. Substitute v = !vrnax and solve for Xl' It is convenientto find X in terms of E and then write E = !kN toobtain an expression for X in terms of A:

1 1 (21T)2 1 ( 21T )2E = -kN = -m - N = -(3.0 kg) - (0.040 m)Z2 2 T 2 2.0 s

= 2.37 X 10-2 J = 12.4 X 10-2 J I!mV~ax = E

so v - f2E_max \j-;; 2(2.37 X 10-2 J) I I

3.0 kg = 0.126 m/s = 0.13 m/s

E = !m(!vrnaY + !kxi = !(!mV~ax) + !kxi = !E + !kxiso !kxi = E - !E = ~E

and X = + f3E = + 1~(.!.kN) = + V3 AI --y 2k --y 2k 2 - 2

= ± v.: (4.0 cm) = I ±3.5 cm I

CHECK As expected, the result for Part (c), step 2 has two values, one with the springextended, the other with the spring compressed. In addition, we expected these values tobe equal, except for the sign. Further, the positive result is less than 4,0 cm (the amplitude is·to cm), as expected.

PRACTICE PROBLEM 14-3 An object of mass 2.00 kg is attached to a spring that has a forceconstant 40.0 N/m. The object is moving at 25.0 cm/s when it is at its equilibrium position.(a) What is the total energy of the object? (b) What is the amplitude of the motion?

Parabola approximating U nearpoint of stable equilibrium

Simple harmonic motion typically occurs when a particle is displaced slightly froma position of stable equilibrium. Figure 14-9 is a graph of the potential energy Uversus x for a force that has a position of stable equilibrium and a position of un-stable equilibrium. As discussed in Chapter 7, the potential-energy maximum at X2on Figure 14-9 corresponds to unstable equilibrium, whereas the minimum at Xl

corresponds to stable equilibrium. Many smooth curves with a minimum as inFigure 14-9 can be closely approximated near the minimum by a parabola. Thedashed curve in this figure is a parabolic curve that approximately fits U near the

FIG U R E 1 4 - 9 Plot of U versus x for a forcethat has a position of stable equilibrium (Xl) anda position of unstable equilibrium (x2).

stable equilibrium point. The general equation for a parabola that has a minimumat point Xl can be written

where A and B are constants. The constant A is the value of U at the equilibriumposition X = Xl' The force is related to the potential energy curve by Ft = -dUjdx.Then

dUFx = - dx = -2B(x - Xl)

If we set 2B = k, this equation reduces to

dUFx = -d; = -k(x - Xl)

According to Equation 14-20, the force is proportional to the displacement fromequilibrium and oppositely directed, so the motion will be simple harmonic.Figure 14-9 shows a graph of this system's potential energy function U(x), whichhas a position of stable equilibrium at X = Xl' Figure 14-10 shows a potential-energy function that has a position of stable equilibrium at X = O.The system forthis function is a small particle of mass m oscillating back and forth at the bottomof a frictionless spherical bowl.

14-3 SOME OSCILLATING SYSTEMS

When an object hangs from a vertical sprrng, there is a downwardforce mg in addition to the force of the spring (Figure 14-11). If wechoose downward as the positive y direction, then the spring's forceon the object is -ky, where y is the extension of the spring. The netforce on the object is then

'LFy = -ky + mg 14-21

We can simplify this equation by changing to a new variabley' = y - yO' where Yo = mgjk is the amount the spring is stretchedwhen the object is in equilibrium. Substituting y' + Yo for y gives

'LFy = -kCy' + Yo) + mg

Position with Fsspringunstretched.

'LFy = -ky'

Newton's second law ('LFy = may) gives

d2y-ky' = m dt2

However, y = y' + yO' where Yo = mgjk is a constant. Thus d2y j dt2 = d2y' j dt2, sod2y'

-ky' = m dt2

which is the same as Equation 14-2 with y' replacing x. It has the now familiarsolution

Actual potenCLenergy functio:-

FIG U R E 1 4 • 1 0 Plot of U versus x for asmall particle oscillating back and forth at thbottom of a spherical bowl.

y

1Equilibrium positionwith mass m attached.Spring stretches anamount Yo = mglk.

Object oscillatesaround the equilibriumposition with a dis-placement y'= y - Yo.

FIG U R E 1 4· 1 1 The Newton's secondlaw equation for the motion of a mass on avertical spring is greatly simplified if thedisplacement 01') is measured from theequilibrium position of the spring with themass attached.

-=nus, the effect of the gravitational force mg is merely to shift the equilibrium-_=ition from y = 0 to y' = O.When the object is displaced from this equilibrium:":-sition by the amount y', the net force is -ky'. The object oscillates about this::-~brium position with an angular frequency w = Yk/m, the same angular fre-:..:ency as that for an object on a horizontal spring .

.-\ force is conservative if the work done by it is independent of the path. Both=-~force of the spring and the force of gravity are conservative, and the sum of=- ~ e forces (Equations 14-21 and 14-22) also is conservative. The potential energy-..:...ction U associated with the sum of these forces is the negative of the work done:":' - an arbitrary integration constant. That is,

U = - I -ky'dy' = !ky'2 + Uo

-~ere the integration constant Uo is the value of U at the equilibrium position0' = 0). Thus,

•• Paper Springs

:0U are showing your nieces how to make paper party decorations using paper springs. One:-..:ecemakes a paper spring. The spring is stretched 8 em and has a single sheet of colored?aper suspended from it. You want the decorations to bounce at approximately 1.0 cy/s.:-low many sheets of colored paper should be used for the decoration on that spring if it is:0 bounce at 1.0 cy/s?

PICTURE The frequency depends on the ratio of the force constant to the suspended massEquation 14-12), and you do not know either the force constant or the mass. However,:looke's law (Equation 14-1) can be used to find the required ratio from the information given.

_. Write the frequency in terms of the force constant kand the mass M (Equation 14-12), where M is themass of N sheets. We need to find N:

The spring stretches a distance of Yo = 8.0 emwhen a single sheet of mass m is suspended:

3. The mass of N sheets equals N times the mass ofa single sheet:

4. Using the step-2 and step-3 results, solve for k/ M:

- Substitute the step-4 result into the step-1 resultand solve for N:

w 1 (Tf = 211"= 211"'Y/Vi

k k 1 gM Nm N Yo

1(T 1{lff = 211"'Y /Vi = 211"'Y NYo

g 9.81 m/s2

N = --- ------- = 31(211"f)Zyo 411"2(1.0Hz)Z(O.080 m) .

I Three sheets are needed. ICHECK Three or more sheets of construction paper seems plausible. Fifty or one-hundredsheets would likely wreck a paper spring.

TAKING IT FURTHER Note that we did not need to use the value of m or k in this exampleecause the frequency depends on the ratio kim, which equals g/yo' In addition, we have ne-

olected the mass of the spring itself. Its mass is probably not negligible compared to the massof a few sheets of construction paper, so our step-5 result is an approximate result.

PRACTICE PROBLEM 14-4 How much is the paper spring stretched when a decoration.•....made from three sheets of paper is suspended from it and the paper is in equilibrium?

A paper spring (under construction).(Rhoda Peacher.)