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Example 14-15 An Object on a Spring

An object of mass 1.5 kg on a spring that has a force constant equal to 600 N/m loses 3.0% ofits energy in each cycle. The same system is driven by a sinusoidal force with a maximumvalue of Fo = 0.50 N. (a) What is Q for this system? (b) What is the resonance (angular) fre-quency? (c) If the driving frequency is slowly varied through resonance, what is the width.:.lW of the resonance? (d) What is the amplitude at resonance? (e) What is the amplitude if thedriving frequency w = 19 rad/s?

PICTURE The energy loss per cycle is only 3.0%, so the damping is weak. We can find Qfrom Q = 27T/(6.E/Elcycle (Equation 14-47) and then use this result and 6.w/wo = l/Q(Equation 14-51) to find the width 6.w of the resonance. The resonance frequency is the nat-ural frequency. The amplitude both at resonance and off resonance can be found fromEquation 14-55, with the damping constant calculated from Q using Q = WOT (Equation 14-45)and T = m/b (Equation 14-42).

SOLVE

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

(a) The damping is weak. Relate Q to the fractional energy lossusing Q = 27T/(6.E/E)cycle (Equation 14-47):

(c) Relate the width of the resonance 6.w to Q using 6.w/wo = l/Q(Equation 14-51):

(d) 1. Write an expression for the amplitude A for any drivingfrequency W (Equation 14-55):

3. Use Q = WOT (Equation 14-45) and T = m/b (Equation 14-42)to relate the damping constant b to Q:

4. Use the results of the previous two steps to calculate theamplitude at resonance:

(e) Calculate the amplitude for W = 19 rad/s. (We omit the units tosimplify the equation. Because all quantities are in 51units,A will be in meters.)

Q = (I6.E~/~)CYcle = ~ = ~

Wo = /!; = 120 rad/s I6.w = ~ = I 0.096 rad/s I

CHECK At a frequency just 1 rad/s below the 20 rad/s resonance frequency, the amplitudedrops by a factor of 20. This is not surprising, because the width 6.w of the resonance is only0.096 rad/s.

TAKING IT FURTHER Off resonance the term b2w2 is negligiblecompared with the other term in the denominator of the expres-sion for A. When W - Wo is more than several times the width 6.w,as it was in this example, we can neglect the b2w2 term and calcu-late A from A = Fo/[m(w5 - w2)]. Figure 14-26 shows the ampli-tude versus driving frequency w. Note that the horizontal scale isover a small range of w.

19.5 20OJ, rad/s

-------------------------------Moving to the Beat: Millennium Bridge

The three-span London Millennium footbridge opened in June of 2000. Between80,000 and 100,000 people crossed the suspension bridge during the course of theday. As the bridge became crowded with up to 2000 people on it at anyone time,*it began to sway from side to side. Soon the lateral swaying was so strong thatmany people had to hold onto the handrails.t The "Wobbly Bridge"+ was closedthree days after its opening and did not reopen until February 2002. The footbridgewas designed to withstand extremely strong winds, as well as hits from heavybarges on both piers. However, the lateral motion was a shock to the designers andengineers. After several months of study, researchers concluded that walking has alateral component of force, as well as vertical and forward/backward components.

The typical cadence of a walking person is such that his or her left foot strikesthe walkway at approximately one-second intervals. The same is true, of course, forthe right foot. When someone steps forward onto his or her left foot, nearly 25 N offorce is directed to the left; the force is directed to the right for the right foot.# Ateach step, a left or right lateral force is exerted on the walkway, so the lateral forcesshake the walkway with a frequency of 1 Hz. Unfortunately, the two lowest naturalfrequencies of sideways motion for the 144-meter-long center span were 0.5 Hz and1.0 Hz,o and the 100-meter-long southern span had a natural vibration mode at 0.8Hz. The footsteps of the crowd drove the motion of the bridge. When the crowdwas small, the combined force of the footsteps was not enough to cause motion. But after more than 200 people were on thebridge,§ the natural damping of the bridge was not high enough to resist the combined force of the crowd's footsteps push-ing the bridge sideways.

The swaying increased because of human reaction to the sideways motion. Calculations show that the maximum lateral ac-celeration was between 0.2g and 0.3g,'ll enough to cause people to lose their balance. An instinctive method of regaining bal-ance on a moving surface is to walk so that the timing of footsteps matches the motion of the surface. This resonant walkingincreased the amplitude of the motion.

Measurements were made of test crowds on the bridge, which led to the solution of a series of dampers. Eight tuned-massdampers and 37 viscous dampers were installed to reduce the lateral swaying. The tuned-mass dampers are 2.5-ton steelblocks suspended on pendulums. They reduce lateral sway by vibrating 1800 out of phase with the bridge.** The viscousdampers are similar to the shock absorbers used to dampen vertical oscillations in automobiles; they work by moving a pis-ton back and forth through a viscous fluid. The main lateral damping is performed by 37 viscous dampers.tt Additional massdampers were installed to dampen any vertical oscillations. During the final tests before reopening, the peak measured ac-celerations on the bridge dropped by 97%, from 0.25g to 0.006g.++ The bridge has had no swaying problems since reopening.

Any##bridge with a lateral vibration mode below 1.3 Hz is susceptible to oscillation caused by the footsteps of a crowd.oo

Several different types of bridges had exhibited lateral swaying under pedestrian loads, including a cable-stayed bridge inJapan§§ and footbridges in Paris and Ottawa. Even highway bridges have shown the same behavior. '11'11 Because of the LondonMillennium footbridge, engineers have been motivated to look at vibration in a new manner.

Massive damped oscillators were attachedunder the walkway shortly after thissuspension bridge opened. The oscillators wereput there to prevent the excessive swaying thatwas driven by lateral forces exerted by thefootsteps of the walkers. (Alamy.)

• Dallard, P., et aI., "The London Millennium Footbridge," The Structural Engineer, Nov. 20, 2001, Vol. 79, o. 22, 17-33.t Smith, Michael, "Bouncing Bridge May Be Closed 'for Weeks:" The Telegraph, Jun. 13, 2000. http://www.telegraph.co.uk/news/main.jhtml?xml=/news/2000/06/13/nsway13.xml

as of July 2006.I Binney, Magnus, "Throwing a Wobbly:' Tile Times, Oct. 31, 2000, Features, 16.• "Oscillation:' The Millennium Bridge - Challenge. Arup Engineering. http://www.arup.com/MillenniumBridge/challenge/ oscillation.hlml as of July 2006.o Fitzpatrick, T., Linking London: The Millennium Bridge. London: The Royal Academy of Engineering, June 200l.§ Roberts, T. M., "Lateral Pedestrian Excitation of Footbridge," Journal of Bridge Engineering, Jan./Feb. 2005, Vol. 10, No.1, 107-112s., Dallard et aJ., op. cit.•• "Elegant, Filigran, and Not Moving:' GERB Vibration Control Systems. http://gerb.com/images/both/projektbeispiele/pdf/millenium_bridge_en.pdf as of July 2006.tt Taylor, D. P., "Damper Retrofit of the London Millenium Footbridge-A Case Study in Biodynamic Design," Taylor Devices. http://www.taylordevices.com/papers/damper/

damper.pdf as of July 2006.II Ibid .•• Strllctural Safety 2000·2001: Thirteenth Report of SCaSS-The Standing Committee on Structural Safety. London: Standing Committee on Structural Safety. May 2001, 24-26.

http://www.scoss.org.uk/publications/rtf/13Report.pdf as of July 2006.00 "Designing Foolbridges with Eurocodes:' Eurocode News, Mar. 2004, No.2, 6.§§ Nakamura, S.-I., "Model for Lateral Excitation of Footbridges by Synchronous Walking," JOllrnalof Strllctural Engineering, Jan. 2004, 32-37." Fitzpatrick, op. cit.

1. Simple harmonic motion occurs whenever the restoring force is proportional to thedisplacement from equilibrium. It has wide application in the study of oscillations, waves,electrical circuits, and molecular dynamics.

2. Resonance is an important phenomenon in many areas of physics. It occurs whenthe frequency of the driving force is close to the natural frequency of the oscillatingsystem.

In simple harmonic motion, the acceleration (and thus the net force) is both proportional to,and oppositely directed from, the displacement from the equilibrium position.

F = -kx = ma 14-1x x

x = A cos(wt + (5)

27TW =27Tf =-

T

If a particle moves in a circle with constant speed, the projection of the particle onto adiameter of the circle moves in simple harmonic motion.

If an object is given a small displacement from a position of stable equilibrium, it typicallyoscillates about this position with simple harmonic motion.

w=~ 14-8

w=Jt 14-27 ;.w=~ 14-33

where I is the moment of inertia and K is the torsional constant. For small oscillations of aphysical pendulum, K = MgD, where D is the distance of the center of mass from therotation axis.

In the oscillations of real systems, the motion is damped because of dissipative forces.If the damping is greater than some critical value, the system does not oscillate whendisturbed, but merely returns to its equilibrium position. The motion of a weakly dampedsystem is nearly simple harmonic with an amplitude that decreases exponentially withtime.

w1=woHE = Eoe-tjT

A = Aor(lj2)tjT

Energy

Amplitude

14-44

14-41

mT =-

b

Q = WOT

( It1EI)-- «1E cycle

When an underdamped (b < be) system is driven by an external sinusoidal forceFex! = Fo cos wt, the system oscillates with a frequency w equal to the driving frequency andan amplitude A that depends on the driving frequency.

Clw 1wa Q

x = A cos(wt - 0)

FaA = -~===========Vm2(w6 - w2)2 + b2w2

bwtanS = (2 2)m Wa - w

Ansvver to Concept Check

14-1 ~

Ansvvers to Practice Problems

14-1 (a) f = 3.6 Hz, T = 0.28 s, (b) f = 2.5 Hz, T = 0.40 s

14-2 w = 3.1 rad/s, vmax = 0.13 m/s

Single-concept, single-step, relatively easy

Intermediate-level, may require synthesis of concepts

Challenging

Solution is in the Student Solutions ManualConsecutive problems that are shaded are pairedproblems.

In a few problems, you are given more data than youactually need; in a few other problems, you are required tosupply data from your general knowledge, outside sources,or informed estimate.

Interpret as significant all digits in numerical values thathave trailing zeros and no decimal points.

14-3

14-4

(a) E = ~mv~ax = 0.0625 J (b) A = Y2Etota/k = 5.59 cm

24cm

g' = 10.3 m/s2, T = 1.96 s

~L

T = - for x = L/6 and for x = L/23g

• True or false:(a) For a simple harmonic oscillator, the period is proportional to

the square of the amplitude.(b) For a simple harmonic oscillator, the frequency does not depend

01.} the amplitude.(c) If the net force on a particle undergoing one-dimensional motion

is proportional to, and oppositely directed from, the displace-ment from equilibrium, the motion is simple harmonic.

2 • If the amplitude of a simple harmonic oscillator is tripled,by what factor is the energy changed?

3 •• An object attached to a spring exhibits simple harmonicmotion with an amplitude of 4.0 cm. When the object is 2.0 cmfrom the equilibrium position, what percentage of its total me-chanical energy is in the form of potential energy? (a) one-quarter,(b) one-third, (c) one-half, (d) two-thirds, (e) three-quarters S

4 •• An object attached to a spring exhibits simple har-monic motion with an amplitude of 10.0 cm. How far from equi-librium will the object be when the system's potential energy isequal to its kinetic energy? (a) 5.00 em, (b) 7.07 cm, (c) 9.00 cm,(d) The distance cannot be determined from the data given.

5 •• Two identical systems each consist of a spring with oneend attached to a block and the other end attached to a wall. Thesprings are horizontal, and the blocks are supported from below bya frictionless horizontal table. The blocks are oscillating in simpleharmonic motions such that the amplitude of the motion of block A

is four times as large as the amplitude of the motion of block B.How do their maximum speeds compare? (a) vAmax = VB max'

(b) vAmax = 2vBmax' (c) V Amax = 4vBmax' (d) This comparison cannotbe done by using the data given.

6 •• Two systems each consist of a spring with one end attachedto a block and the other end attached to a wall. The springs are hori-zontal, and the blocks are supported from below by a frictionless hor-izontal table. The identical blocks are oscillating in simple harmonicmotions with equal amplitudes. However, the force constant of springA is four times as large as the force constant of spring B. How do theirmaximum speeds compare? (a) v Amax = vBmax' (b) v Amax = 2vBmax'

(c) VAmax = 4vBmax' (d) This comparison cannot be done by usingthe data given.

7 •• Two systems each consist of a spring with one endattached to a block and the other end attached to a wall. Theidentical springs are horizontal, and the blocks are supportedfrom below by a frictionless horizontal table. The blocks areoscillating in simple harmonic motions with equal amplitudes.However, the mass of block A is four times as large as themass of block B. How do their maximum speeds compare?(a) v Amax = vBmax' (b) v Amax = 2vBmax' (c) VAmax = ~vBmax' (d) Thiscomparison cannot be done by using the data given. SSM

8 •• Two systems each consist of a spring with one endattached to a block and the other end attached to a wall. Theidentical springs are horizontal, and the blocks are supportedfrom below by a frictionless horizontal table. The blocks areoscillating in simple harmonic motions with equal amplitudes.However, the mass of block A is four times as large as the

tlw 1wa Q

x = A cos(wt - 8)

FaA = -;::============Vm2(w6 - w2)Z + b2w2

bwtan8 = (2 2)m wa - w

Ansvver to Concept Check

14-1 ~

Ansvvers to Practice Problems

14-1 (a) f = 3.6 Hz, T = 0.28 s, (b) f = 2.5 Hz, T = 0.40 s

14-2 w = 3.1 rad/s, vmax = 0.13 m/s

Single-concept, single-step, relatively easy

Intermediate-level, may require synthesis of concepts

Challenging

Solution is in the Student Solutions ManualConsecutive problems that are shaded are pairedproblems.

In a few problems, you are given more data than youactually need; in a few other problems, you are required tosupply data from your general knowledge, outside sources,or informed estimate.

Interpret as significant all digits in numerical values thathave trailing zeros and no decimal points.

14-3

14-4

(a) E = imv~ax = 0.0625 J (b) A = V2Etota/k = 5.59 em

24 em

g' = 10.3 m/s2, T = 1.96 s

~L

T = - for x = L/6 and for x = L/23g

• True or false:(a) For a simple harmonic oscillator, the period is proportional to

the square of the amplitude.(b) For a simple harmonic oscillator, the frequency does not depend

oJ:}the amplitude.(c) If the net force on a particle undergoing one-dimensional motion

is proportional to, and oppositely directed from, the displace-ment from equilibrium, the motion is simple harmonic.

2 • If the amplitude of a simple harmonic oscillator is tripled,by what factor is the energy changed?

3 •• An object attached to a spring exhibits simple harmonicmotion with an amplitude of 4.0 em. When the object is 2.0 emfrom the equilibrium position, what percentage of its total me-chanical energy is in the form of potential energy? (a) one-quarter,(b) one-third, (c) one-half, (d) two-thirds, (e) three-quarters s

4 •• An object attached to a spring exhibits simple har-monic motion with an amplitude of 10.0 em. How far from equi-librium will the object be when the system's potential energy isequal to its kinetic energy? (a) 5.00 em, (b) 7.07 em, (c) 9.00 em,(d) The distance cannot be determined' from the data given.

5 •• Two identical systems each consist of a spring with oneend attached to a block and the other end attached to a wall. Thesprings are horizontal, and the blocks are supported from below bya frictionless horizontal table. The blocks are oscillating in simpleharmonic motions such that the amplitude of the motion of block A

is four times as large as the amplitude of the motion of block B.How do their maximum speeds compare? (a) vAmax = VB max'

(b) v Amax = 2vBmax' (c) V Amax = 4vBmax' (d) This comparison cannotbe done by using the data given.

6 •• Two systems each consist of a spring with one end attachedto a block and the other end attached to a wall. The springs are hori-zontal, and the blocks are supported from below by a frictionless hor-izontal table. The identical blocks are oscillating in simple harmonicmotions with equal amplitudes. However, the force constant of springA is four times as large as the force constant of spring B. How do theirmaximum speeds compare? (a) v Amax = vBmax' (b) v Amax = 2vBmax'(c) VAmax = 4vBmax' (d) This comparison cannot be done by usingthe data given.

7 •• Two systems each consist of a spring with one endattached to a block and the other end attached to a wall. Theidentical springs are horizontal, and the blocks are supportedfrom below by a frictionless horizontal table. The blocks areoscillating in simple harmonic motions with equal amplitudes.However, the mass of block A is four times as large as themass of block B. How do their maximum speeds compare?(a) vAmax = vBmax' (b) vAmax = 2vBmax' (c) VAmax = ivBmax' (d) Thiscomparison cannot be done by using the data given. s

8 •• Two systems each consist of a spring with one endattached to a block and the other end attached to a wall. Theidentical springs are horizontal, and the blocks are supportedfrom below by a frictionless horizontal table. The blocks areoscillating in simple harmonic motions with equal amplitudes.However, the mass of block A is four times as large as the

: block B. How do the magnitudes of their maximum=rations compare? (a) aAmax = aBmax' (b) aAmax = 2aBma~'

= !aBmax' (d) aAmax = !aBmax' (e) This comparison cannote by using the data given.

•• in general physics courses, the mass of the spring in sim-.onic motion is usually neglected because its mass is usu-

smaller than the mass of the object attached to it._ e:::,this is not always the case. If you neglect the mass of the

. 'hen it is not negligible, how will your calculation of the- 5 period, frequency, and total energy compare to the actual

: these parameters? Explain. SSM

•• Two mass-spring systems oscillate with periods TA and Ts'= :Ta and the systems' springs have identical force constants, it__ at the systems' masses are related by (a) mA = 4mB,= mBlVi, (c) mA = mB/2, (d) mA = ms/4.•• Two mass-spring systems oscillate at frequencies fA and fB'

= -.~aand the systems' springs have identical force constants, it_ that the systems' masses are related by (a) mA = 4mB,= mBlVi, (c) mA = mB/2, (d) mA = ms/4.•• Two mass-spring systems A and B oscillate so that their

::;cchanical energies are equal. If mA = 2mB, which expression-='.3. es their amplitudes? (a) AA = AB/4, (b) AA = ABlVi,

= AB, (d) Not enough information is given to determine the_ :he amplitudes.•• Two mass-spring systems A and B oscillate so that their

anical energies are equal. If the force constant of spring A=:TIesas large as the force constant of spring B, then which

-=.5lon best relates their amplitudes? (a) AA = AB/4,= AB lVi, (c) AA = AB, (d) Not enough information is given

~==u'n<e the ratio of the amplitudes. SSM

•• The length of the string or wire supporting a pendulum--eases slightly when the temperature of the string or wire in-

How does this affect a clock operated by a simple pendulum?,. A lamp hanging from the ceiling of the club car in a train

- with period To when the train is at rest. The period will be:eft and right columns)

than Ta when A. The train moves horizontally atconstant velocity.

=- To when B. The train rounds a curve at constantspeed.

~ :0 To when C. The train climbs a hill at constantspeed.

D. The train goes over the crest of ahill at constant speed.

• Two simple pendulums are related as follows. Pendulum A.!rth LA and a bob of mass mA; pendulum B has a length LB

_ b of mass mB' If the period of A is twice that of B,. L" = 2Ls and mA = 2mB, (b) LA = 4LB and mA = ms'

- = u.B, whatever the ratio mAlmB' (d) LA = ViLB, whatevermjmB·

•• Two simple pendulums are related as follows.- '.lffi A has a length LA and a bob of mass mA; pendulum B

;ength LB and a bob of mass mB• If the frequency of A is-.:... the frequency of B, then (a) LA = 3LB and mA = 3mB,

- = 9LB and mA = mB' (c) LA = 9LB, regardless of the._~mB' (d) LA = \l3LB regardless of the ratio mAlmB' M

•• Two simple pendulums are related as follows.- :.ID1. A has a length LA and a bob of mass mA; pendulum B

!rth L B and a bob of mass mB. They have the same period.:ly difference between their motions is that the amplitude of:ion is twice the amplitude of B's motion, then (a) LA = LB

= mB' (b) LA = 2LB and mA = mB' (c) LA = LB whatevermAimS' (d) LA = !LB· whatever the ratio mAlmB'

19 •• True or false:(a) The mechanical energy of a damped, undriven oscillator de-

creases exponentially with time.(b) Resonance for a damped, driven oscillator occurs when the dri-

ving frequency exactly equals the natural frequency.(c) If the Q factor of a damped oscillator is high, then its resonance

curve will be narrow .(d) The decay time T for a spring-mass oscillator with linear damp-

ing is independent of its mass.(e) The Q factor for a driven spring-mass oscillator with linear

damping is independent of its mass.

20 •• Two damped spring-mass oscillating systems have iden-tical spring and damping constants. However, system A's mass m A

is four times system B's mass mB' How do their decay times com-pare? (a) T A = 4T 8' (b) T A = 2T B' (c) TA = TB' (d) Their decay timescannot be compared, given the information provided.

21 •• Two damped spring-mass oscillating systems have iden-tical spring constants and decay times. However, system A's massmA is twice system B's mass mB' How do their damping constants,b, compare? (a) bA = 4bB, (b) bA = 2bB, (c) bA = bB' (d) bA = !bs'(e) Their decay times cannot be compared, given the informationprovided.

22 •• Two damped, driven spring-mass oscillating systemshave identical driving forces as well as identical spring and damp-ing constants. However, the mass of system A is four times the massof system B. Assume both systems are very weakly damped. Howdo their resonant frequencies compare? (a) wA = wB (b) wA = 2wB,(c) wA = !wB' (d) wA = !wB' (e) Their resonant frequencies cannot becompared, given the information provided.

23 •• Two damped, driven spring-mass oscillating sys-tems have identical masses, driving forces, and damping con-stants. However, system A's force constant kA is four times sys-tem B's force constant kB• Assume they are both very weaklydamped. How do their resonant frequencies compare?(a) wA = WB' (b) wA = 2wB, (c) wA = !wB' (d) wA = !wB' (e) Theirresonant frequencies cannot be compared, given the informa-tion provided. SM

24 •• Two damped, driven simple-pendulum systems haveidentical masses, driving forces, and damping constants. However,system A's length is four times system B's length. Assume they areboth very weakly damped. How do their resonant frequenciescompare? (a) wA = wB' (b) wA = 2wB, (c) wA = !wB' (d) wA = !wB'(e) Their resonant frequencies cannot be compared, given the infor-mation provided.

25 • Estimate the width of a typical grandfather clock's cabi-net relative to the width of the pendulum bob, presuming thedesired motion of the pendulum is simple harmonic. SSM

26 • A small punching bag for boxing workouts is approxi-mately the size and weight of a person's head and is suspendedfrom a very short rope or chain. Estimate the natural frequency ofoscillations of such a punching bag .

27 •• For a child on a swing, the amplitude drops by a factor oflie in about eight periods if no additional mechanical energy.isgiven to the system. Estimate the Q factor for this system.

28 •• (a) Estimate the natural period of oscillation for swingingyour arms as you walk, when your hands are empty. (b) Now esti-mate the natural period of oscillation when you are carrying aheavy briefcase. (c) Observe other people while they walk. Do yourestimates seem reasonable?

:\-ote: Unless otherWise specified, assume that all objects inthis section are in simple harmonic motion.

29 • The position of a particle is given by x = (7.0 cm)cos67T1,where I is in seconds. What are (a) the frequency, (b) the period, and(c) the amplitude of the particle's motion? (d) What is the first timeafter I = 0 that the particle is at its equilibrium position? In what di-rection is it moving at that time?

30 • What is the phase constant 0 in x = A cos (wi + 0)(Equation 14.4) if the position of the oscillating particle at time I = 0is (a) 0, (b) - A, (c) A, and (d) Aj2?

31 • A particle of mass m begins at rest from x = +25 cm andoscillates about its equilibrium position at x = 0 with a period of1.5 s. Write expressions for (a) the position x as a function of I,(b) the velocity Vx as a function of I, and (c) the acceleration ax as afunction of I. SSM

32 •• Find (a) the maximum speed, and (b) the maximum ac-celeration of the particle in Problem 29. (c) What is the first time thatthe particle is at x = 0 and moving to the right?33 •• Work Problem 31 for when the particle is initially atx = 25 cm and moving with velocity Vo = +50 cm/s.

34 •• The period of a particle that is oscillating in simple har-monic motion is 8.0 s and its amplitude is 12 em. At I = 0, it is at itsequilibrium position. Find the distance the particle travels duringthe intervals (a) I = 0 to I = 2.0 s, (b) I = 2.0 s to I = 4.0 s, (c) t = 0to I = 1.0 s, and (d) t = 1.0 s to t = 2.0 S.

35 •• The period of a particle oscillating in simple harmonicmotion is 8.0 s. At t = 0, the particle is at rest at x = A = 10 em.(a) Sketch x as a function of I. (b) Find the distance traveled in thefirst, second, third, and fourth second after I = O.

36 •• ENGINEERING ApPLICATION, CONTEXT-RICH Militaryspecifications often call for electronic devices to be able to with-stand accelerations of up to 109 (lOg = 98.1 m/s2). To make surethat your company's products meet this specification, your man-ager has told you to use a "shaking table," which can vibrate a de-vice at controlled and adjustable frequencies and amplitudes. If adevice is placed on the table and made to oscillate at an amplitudeof 1.5 em, what should you adjust the frequency to in order to testfor compliance with the 109 military specification?

37 •• The position of a particle is given by x = 2.5 cos 7Tt,where x is in meters and t is in seconds. (a) Find the maximumspeed and maximum acceleration of the particle. (b) Find the speedand acceleration of the particle when x = 1.5 m. SSM

38 ••• (a) Show that Ao cos(wt + 0) can be written asAs sin(wt) + Ac cos(wt), and determine As and Ac in terms of Aoand O. (b) Relate Ac and As to the initial position and velocity of aparticle undergoing simple harmonic motion.

SIMPLE HARMONIC MOTIONAS RELATED TO CIRCULAR MOTION

39 • A particle moves at a constant speed of 80 cm/s in acircle of radius 40 em centered at the origin. (a) Find the fre-quency and period of the x component of its position. (b) Writean expression for the x component of the particle's position as afunction of time t, assuming that the particle is located on the +yaxis at time I = O. SSM

40 • A particle moves in a IS-em-radius circle centered atthe origin and completes 1.0 rev every 3.0 s. (a) Find the speedof the particle. (b) Find its angular speed w. (c) Write an equationfor the x component of the particle's position as a function oftime I, assuming that the particle is on the -x axis at time I = O.

ENERGY IN SIMPLEHARMONIC MOTION

41 • A 2.4-kg object on a frictionless hoizontal surface is at-tached to one end of a horizontal spring of force constantk = 4.5 kN/m. The other end of the spring is held stationary. Thespring is stretched 10 em from equilibrium and released. Find thesystem's total mechanical energy.42 • Find the total energy of a system consisting of a 3.0-kgobject on a frictionless horizontal surface oscillating with an amplitudeof 10 em and a frequency of 2.4 Hz at the end of a horizontal spring.

43 • A 1.50-kg object on a frictionless horizontal surface os-cillates at the end of a spring (force constant k = 500 N/m). The ob-ject's maximum speed is 70.0 cm/s. (a) What is the system's totalmechanical energy? (b) What is the amplitude of the motion? SM

44 • A 3.0-kg object on a frictionless horizontal surface os-cillating at the end of a spring that has a force constant equal to2.0 kN/m has a total mechanical energy of 0.90 J. (a) What is theamplitude of the motion? (b) What is the maximum speed?

45 • An object on a frictionless horizontal surface oscillates at theend of a spring with an amplitude of 4.5 em. Its total mechanical en-ergy is 1.4 J. What is the force constant of the spring?46 •• A 3.0-kg object on a frictionless horizontal surfac~ oscil-lates at the end of a spring with an amplitude of 8.0 em. Its maxi-mum acceleration is 3.5 m/s2. Find the total mechanical energy.

SIMPLE HARMONIC MOTIONAND SPRINGS

47 • A 2.4-kg object on a frictionless horizontal surface is at-tached to the end of a horizontal spring that has a force constantk = 4.5 kN/m. The spring is stretched 10 cm from equilibrium andreleased. What are (a) the frequency of the motion, (b) the period,(c) the amplitude, (d) the maximum speed, and (e) the maximum ac-celeration? (j) When does the object first reach its equilibrium posi-tion? What is its acceleration at this time?48 • A 5.00-kg object on a frictionless horizontal surface is at-tached to one end of a horizontal spring that has a force constantk = 700 N/m. The spring is stretched 8.00 em from equilibrium andreleased. What are (a) the frequency of the motion, (b) the period, (c)the amplitude, (d) the maximum speed, and (e) the maximum ac-celeration? (j) When does the object first reach its equilibrium posi-tion? What is its acceleration at this time?49 • A 3.0-kg object on a frictionless horizonal surface is at-tached to one end of a horizontal spring and oscillates with an am-plitude A = 10 cm and a frequency f = 2.4 Hz. (a) What is the forceconstant of the spring? (b) What is the period of the motion?(c) What is the maximum speed of the object? (d) What is the max-imum acceleration of the object? ss50 • An 85.0-kg person steps into a car of mass 2400 kg, caus-ing it to sink 2.35 em on its springs. If started into vertical oscilla-tion, and assuming no damping, at what frequency will the car andpassenger vibrate on these springs?51 • A 4.50-kg object oscillates on a horizontal spring with anamplitude of 3.80 cm. The object's maximum acceleration is 26.0 m/s2•

Find (a) the force constant of the spring, (b) the frequency of the object,and (c) the period of the motion of the object.52 •• An object of mass m is suspended from a vertical springof force constant 1800 N/m. When the object is pulled down 2.50 emfrom equilibrium and released from rest, the object oscillates at5.50 Hz. (a) Find m. (b) Find the amount the spring is stretched fromits unstressed length when the object is in equilibrium. (c) Writeexpressions for the displacement x, the velocity Vx and the acceler-ation ax as functions of time I.

::: •• An object is hung on the end of a vertical spring and is::=:eased from rest with the spring unstressed. If the object falls:L em before first coming to rest, find the period of the resulting=illatory motion.

•• A suitcase of mass~- g is hung from two bungee__ ,as shown in Figure 14-27.~ cord is stretched 5.0 em when=-.£ suitcase is in equilibrium. If the::--tease is pulled down a little and

ed, what will be its oscillation=-aJuency?:5 •• A O.l20-kg block is sus-?='1ded from a spring. When a=all pebble of mass 30 g is placed__, the block, the spring stretches an--' "tional 5.0 em. With the pebble

, the block, the block oscillatesc-::han amplitude of 12 em. (a) What is the frequency of the motion?:: How long does the block take to travel from its lowest point to its_ est point? (c) What is the net force on the pebble when it is at the

: . t of maximum upward displacement?•• Referring to Problem 55, find the maximum amplitude of

::s:illation at which the pebble will remain in contact with the block.

•• An object of mass 2.0 kg is attached to the top of a verti-~ spring that is anchored to the floor. The unstressed length of the~g is 8.0 em and the length of the spring when the object is in=--:-:.illibriumis 5.0 em. When the object is resting at its equilibrium_ s:ition, it is given a sharp downward blow with a hammer so that~ initial speed is 0.30 m/s. (a) To what maximum height above:::.e floor does the object eventually rise? (b) How long does it take

_ the object to reach its maximum height for the first time?: Does the spring ever become unstressed? What minimum initial"':ocity must be given to the object for the spring to be unstressed

-: me time?

FIGURE 14-27

Problem 54

Oil • •• ENGINEERING ApPLICATION A winch cable has a cross-sa::ional area of 1.5 cm2 and a length of 2.5 m. Young's modulus for:::.e cable is 150 GN/m2. A 950-kg engine block is hung from the end

- , e cable. (a) By what length does the cable stretch? (b) If we treat:::.e cable as a simple spring, what is the oscillation frequency of the

:' e block at the end of the cable?

• Find the length of a simple pendulum if its frequencysmall amplitudes is 0.75 Hz. M

• Find the length of a simple pendulum if its period for;;;nall amplitudes is 5.0 s.

• What would the period of the pendulum in Problem 60.:E 'i the pendulum were on the moon, where the acceleration due

yavity is one-sixth that on Earth?

• If the period of a 70.0-cm-Iong simple pendulum is 1.68 s,_ is the value of g at the location of the pendulum?

• A simple pendulum that is set up in the stairwell of a 10-__: building consists of a heavy weight suspended on a 34.0-m-:: wire. What is the period of oscillation?

•• Show that the total energy of a simple pendulum under-::-_~ oscillations of small amplitude 4>0 (in radians) is E = ~mgL4>6'-: _;:Use the approximation cos 4> = 1 - ~4>2 for small 4>.

• •• A simple pendulum of length L is attached to a massive_ that slides without friction down a plane inclined at angle fJ

..:. the horizontal, as shown in Figure 14-28. Find the period of::s::llation for small oscillations of this pendulum. SSM

66 ••• The bob at the end of a simple pendulum of length L isreleased from rest from an angle 4>0- (a) Model the pendulum'smotion as simple harmonic motion, and find its speed as it passesthrough 4> = 0 by using the small angle approximation. (b) Usingthe conservation of energy, find this speed exactly for any angle(not just small angles). (c) Show that your result from Part (b)agrees with the approximate answer in Part (a) when 4>0 is small.(d) Find the difference between the approximate and exact resultsfor 4>0 = 0.20 rad and L = 1.0 m. (e) Find the difference betweenthe approximate and exact results for 4>0 = 1.20 rad andL = 1.0m.

67 • A thin 5.0-kg uniform disk with a 20-cm radius is free torotate about a fixed horizontal axis perpendicular to the disk andpassing through its rim. The disk is displaced slightly from equilib-rium and released. Find the period of the subsequent simple har-monic motion. SSM

68 • A circular hoop that has a 50-em radius is hung on a narrowhorizontal rod and allowed to swing in the plane of the hoop. What isthe period of its oscillation, assuming that the amplitude is small?

69 • A 3.0-kg plane figure is suspended at a point 10 emfrom its center of mass. When it is oscillating with small ampli-tude, the period of oscillation is 2.6 s. Find the moment of inertiaI about an axis perpendicular to the plane of the figure throughthe pivot point.

70 •• ENGINEERING ApPLICATION, CONTEXT-RICH, CONCEPTUAL

You have designed a cat door that consists of a square piece of ply-wood that is 1.0 in. thick and 6.0 in. on a side, and is hinged at itstop. To make sure the cat has enough time to get through it safely,the door should have a natural period of at least 1.0 s. Will your de-sign work? If not, explain qualitatively what you would need to doto make it meet your requirements.

71 •• You are given a meterstick and asked to drill a small di-ameter hole through it so that, when the stick is pivoted about ahorizontal axis through the hole, the period of the pendulum willbe a minimum. Where should you drill the hole?

72 •• Figure 14-29 shows a uniformdisk that has a radius R = 0.80 m, a massof 6.00 kg, and a small hole a distance dfrom the disk's center that can serve asa pivot point. (a) What should be thedistance d so that the period of thisphysical pendulum is 2.50 s? (b) Whatshould be the distance d so that thisphysical pendulum will have the shortestpossible period? What is this shortest pos-sible period?

FIGURE 14-29

Problem 72

73 • •• Points PI and Pz on a planeobject (Figure 14-30), are distances hIand hz respectively, from the center ofmass. The object oscillates with thesame period T when it is free to rotateabout an axis through PI and when it isfree to rotate about an axis through Pz.Both of these axes are perpendicular tothe plane of the object. Show thathI + hz = gTz/(47TZ), where hI *" hz·

SSM

74 • •• A physical pendulum con-sists of a spherical bob of radius r andmass m suspended from a rigid rod ofnegligible mass, as in Figure 14-31. Thedistance from the center of the sphereto the point of support is L. When r ismuch less than L, such a pendulumis often treated as a simple pendu-lum of length L. (a) Show that theperiod for small oscillations is givenby T = T VI + (2rz/5U), whereTo = 27Tvijg is the period of a simplependulum of length L. (b) Show thatwhen r is smaller than L, theperiod can be approximated byT = To(1 + rZ/5U). (c) If L = 1.00 mand r = 2.00 cm, find the error in thecalculated value when the approxima-tion T = To is used for this period.How large must the radius of the bobbe for the error to be 1.00 percent?75 • •• Figure 14-32shows the pendulum of a clock in your grand-mother's house. The uniform rod of length L = 2.00 m has a massm = 0.800 kg. Attached to the rod is a uniform disk of massM = 1.20 kg and radius 0.150 m. The clock is constructed to keep per-fect time if the period of the pendulum is exactly 3.50 s. (a) Whatshould the distance d be so that the period of this pendulum is 2.50 s?(b) Suppose that the pendulum clock loses 5.00 min/d. To make sureyour grandmother will not be late for her quilting parties, you decideto adjust the clock back to its proper period. How far and in what di-rection should you move the disk to ensure that the clock will keepperfect time?

FIGURE 14-30

Problem 73

FIGURE 14-31

Problem 74

TT m

FIGURE 14-32

Problem 75

76 • A 2.00-kg object oscillates on a spring with an initialamplitude of 3.00 cm. The force constant of the spring is 400 N/m.Find (a) the period, and (b) the total initial energy. (c) If the energydecreases by 1.00 percent per period, find the linear damping con-stant b and the Q factor.

77 •• Show that the ratio of the amplitudes for two succes-sive oscillations is constant for a linearly damped oscillator. SSM

78 •• An oscillator has a period of 3.00 s. Its amplitude de-creases by 5.00 percent during each cycle. (a) By how much doesits mechanical energy decrease during each cycle? (b) What isthe time constant T? (c) What is the Q factor?

79 •• A linearly damped oscillator has a Q factor of 20. (a) Bywhat fraction does the energy decrease during each cycle? (b) UseEquation 14-40 to r~d the percentage difference between w' and wOoHint: Use the approximation (1 + X)I/Z = 1 + ~x for small x.

80 •• A linearly damped mass-spring system oscillates at200 Hz. The time constant of the system is 2.0 s. At t = 0, the am-plitude of oscillation is 6.0 cm and the energy of the oscillating sys-tem is 60 J. (a) What are the amplitudes of oscillation at t = 2.0 sand t = 4.0 s? (b) How much energy is dissipated in the first 2-s in-terval and in the second 2-s interval?

81 •• ENGINEERING ApPLICATION Seismologists and geophysi-cists have determined that the vibrating Earth has a resonance pe-riod of 54 min and a Q factor of about 400. After a large earthquake,Earth will "ring" (continue to vibrate) for up to 2 months. (a) Findthe percentage of the energy of vibration lost to damping forcesduring each cycle. (b) Show that after n periods the vibrational en-ergy is given by En = (0.984)" Eo, where Eo, is the original energy.(c) If the original energy of vibration of an earthquake is Eo, what isthe energy after 2.0 d? SS

82 ••• A pendulum that is used in your physics laboratory ex-periment has a length of 75 cm and a compact bob with a massequal to 15 g. To start the bob oscillating, you place a fan next to itthat blows a horizontal stream of air on the bob. While the fan is on,the bob is in equilibrium when the pendulum is displaced by anangle of 5.0° from the vertical. The speed of the air from the fan is7.0 m/s. You turn the fan off, and allow the pendulum to oscillate.(a) Assuming that the drag force due to the air is of the form -bv,predict the decay time constant T for this pendulum. (b) How longwill it take for the pendulum's amplitude to reach 1.00?

83 ••• ENGINEERING ApPLICATION, CONTEXT-RICH You are incharge of monitoring the viscosity of oils at a manufacturingplant and you determine the viscosity of an oil by using the fol-lowing method: The viscosity of a fluid can be measured by de-termining the decay time of oscillations for an oscillator that hasknown properties and operates while immersed in the fluid. Aslong as the speed of the oscillator through the fluid is relativelysmall, so that turbulence is not a factor, the drag force of the fluidon a sphere is proportional to the sphere's speed v relative to thefluid: Fd = 67Ta7)V, where 7) is the viscosity of the fluid and a is thesphere's radius. Thus, the constant b is given by 67Ta7). Supposeyour apparatus consists of a stiff spring that has a force constantequal to 350 N/cm and a gold sphere (radius 6.00 cm) hanging onthe spring. (a) What viscosity of an oil do you measure if thedecay time for this system is 2.80 s? (b) What is the Q factor foryour system? SSM

DRIVEN OSCILLATIONSAND RESONANCE

84 • A linearly damped oscillator loses 2.00 percent of its en-ergy during each cycle. (a) What is its Q factor? (b) If its resonancefrequency is 300 Hz, what is the width of the resonance curve Llwwhen the oscillator is driven?

85 • Find the resonance frequency for each of the three systemsshown in Figure 14-33.

:9«1k = 800.0N/m .~

9:

86 •• A damped oscillator loses 3.50 percent of its energy dur-ing each cycle. (a) How many cycles elapse before half of its origi-nal energy is dissipated? (b) What is its Q factor? (c) If the naturalfrequency is 100 Hz, what is the width of the resonance curve whenthe oscillator is driven by a sinusoidal force?

87 •• A 2.00-kg object oscillates on a spring that has a forceconstant equal to 400 N/m. The linear damping constant has avalue of b = 2.00 kg/so The system is driven by a sinusoidalforce of maximum value 10.0 N and angular frequencyw = 10.0 rad/s. (a) What is the amplitude of the oscillations?(b) If the driving frequency is varied, at what frequency will res-onance occur? (c) What is the amplitude of oscillation at reso-nance? (d) What is the width of the resonance curve t1w? SSM

88 •• ENGINEERING ApPLICATION, CONTEXT-RICH Supposeyou have the same apparatus that is described in Problem 83 andthe same gold sphere hanging from a weaker spring that has aforce constant of only 35.0 N/cm. You have studied the viscosityof ethylene glycol with this device, and found that ethylene glycolhas a viscosity value of 19.9 mPa' s. Now you decide to drive thissystem with an external oscillating force. (a) lf the magnitude ofthe driving force for the device is 0.110N and the device is drivenat resonance, how large would be the amplitude of the resultingoscillation? (b) If the system were not driven, but were allowed tooscillate, what percentage of its energy would it lose per cycle?

89 • MULTISTEP A particle's displacement from equilibriumis given by x(t) = 0.40 cos(3.0t + 7T/4), where x is in meters and t isin seconds. (a) Find the frequency f and period T of its motion.(b) Find an expression for the velocity of the particle as a functionof time. (c) What is its maximum speed?90 • ENGINEERING ApPLICATION An astronaut arrives at anew planet, and gets out his simple device to determine the gravi-tational acceleration there. Prior to his arrival, he noted that the ra-dius of the planet was 7550 km. If his 0.500-m-long simple pendu-lum has a period of 1.0 s, what is the mass of the planet?91 •• A pendulum clock keeps perfect time on Earth's surface.In which case will the error be greater: if the clock is placed in amine of depth h, or if the clock is elevated to a height h? Prove youranswer and assume that h « RE.

92 •• Figure 14-34 shows a pendulum of length L with a bob ofmass M. The bob is attached to a spring that has a force constant k,as shown. When the bob is directly below the pendulum support,the spring is unstressed. (a) Derive an expression for the period of

this oscillating system for small-amplitude vibrations. (b) Supposethat M = 1.00 kg and L is suchthat in the absence of the springthe period is 2.00 s. What is theforce constant k if the period of theoscillating system is 1.00 s?93 •• A block that has a massequal to mj is supported frombelow by a frictionless horizontalsurface. The block, which is at-tached to the end of a horizontalspring that has a force constant k,oscillates with an amplitude A.When the spring is at its greatestextension and the block is instantaneously at rest, a second block ofmass m2 is placed on top of it. (a) What is the smallest value for thecoefficient of stati friction P-s such that the second object does notslip on the first? (b) Explain how the total mechanical energy E, theamplitude A, the angular frequency w, and the period T of thesystem are affected by the placing of m2 on mj, assuming that the co-efficient of friction is great enough to prevent slippage. SSM

94 •• A 100-kg box hangs from the ceiling of a room-suspended from a spring with a force constant of 500 N/m. The un-stressed length of the spring is 0.500 m. (a) Find the equilibrium po-sition of the box. (b) An identical spring is stretched and attached tothe ceiling and the box, and is parallel with the first spring. Find thefrequency of the oscillations when the box is released. (c) What isthe new equilibrium position of the box once it comes to rest?95 •• ENGINEERING ApPLICATION The acceleration due togravity g varies with geographical location because of Earth'srotation and because Earth is not exactly spherical. This was firstdiscovered in the seventeenth century, when it was noted that apendulum clock carefully adjusted to keep correct time in Paris lostabout 90 sid near the equator. (a) Show by using the differential ap-proximation that a small change in the acceleration of gravity t1gproduces a small change in the period t1T of a pendulum given byt1T/T = -!t1g/g. (b) How large a change in g is needed to accountfor a 90-s/d change in the period?96 •• A small block that has a mass equal to mj rests on a pis-ton that is vibrating vertically with simple harmonic motion de-scribed by the formula y = A sin wt. (a) Show that the block willleave the piston if w2A > g. (b) If w2 A = 3g and A = 15 cm, at whattime will the block leave the piston?97 •• Show that for the situations in Figure 14-35a and 14-35b,the object oscillates with a frequency f = (1/27T)Vkeff/m, where keffis given by (a) keff = kj + k2, and (b) l/keff = l/kj + 1/k2• Hint: Findthe magnitude of the net force F on the object for a small displacement xand write F = -keffX. Note that in Part(b) the springs stretch by differ-ent amounts, the sum of which is X. SSM

FIGURE 14-34

Problem 92

FIG U RE 14 - 3 5 Problem 97

98 •• CONTEXT-RICH During an earthquake, a horizontal flooroscillates horizontally in approximately simple harmonic motion.Assume it oscillates at a single frequency with a period of 0.80 s.(a) After the earthquake, you are in charge of examining the video ofthe floor motion and discover that a box on the floor started to slip

when the amplitude reached 10 em. From your data, determine thecoefficient of static friction between the box and the floor (b) If the co-efficient of friction between the box and floor were 0040, what wouldbe the maximum amplitude of vibration before the box would slip?99 •• If we attach two blocks that have masses ml and m2 toeither end of a spring that has a force constant k and set them intooscillation by releasing them from rest with the spring stretched,show that the oscillation frequency is given by w = (k/ J-L)1/2,whereJ-L = mlm2/(ml + m2) is the reduced mass of the system.100 •• In one of your chemistry labs, you determine that oneof the vibrational modes of the HCI molecule has a frequency of8.969 X 1013 Hz. Using the result of Problem 99, find the "effectivespring constant" between the H atom and the CI atom in the HClmolecule.101 •• If a hydrogen atom in HCl were replaced by a deuteriumatom (forming DCI) in Problem 100, what would be the newvibration frequency of the molecule? Deuterium consists of 1 protonand 1 neutron.102 ••• SPREADSH EET Ablock of mass m resting on ahorizontal table is attachedto a spring that has a forceconstant k, as shown inFigure 14--36.The coefficientof kinetic friction between the block and the table is J-Lk' The spring isunstressed if the block is at the origin (x = 0), and the +x direction isto the right. The spring is stretched a distance A, where kA > ~mg,and the block is released. (a) Apply Newton's second law to the blockto obtain an equation for its acceleration d2x/dt2 for the first half-cycle, during which the block is moving to the left. Show that the re-sulting~uation can be written as d2x'/dt2 = -w2x', wherew = Vf7m and x' = x - XO' with Xo = J-Lkmg/k = J-Lkg/W2 (b) RepeatPart (a) for the second half-eycle as the block moves to the right, andshow that d2x'/dt2 = -w2x', where x' = x + Xo and Xo has the samevalue. (e) Use a spreadsheet program to graph the first five half-cy-cles for A = lOxo' Describe the motion, if any, after the fifth half-eycle.103 ••• Figure 14--37 shows a uni-form solid half-eylinder of mass M andradius R resting on a horizontal surface.If one side of this cylinder is pusheddown slightly and then released, thehalf-cylinder will oscillate about itsequilibrium position. Determine the pe-- FIG U R E 1 4 - 3 7

riod of this oscillation. SSM Problem 103104 • • • A straight tunnel isdug through Earth, as shownin Figure 14-38. Assume thatthe walls of the tunnel are fric-tionless. (a) The gravitationalforce exerted by Earth on a par-ticle of mass m at a distance rfrom the center of Earth whenr < RE is Fr = -(GmME/R~)r,where ME is the mass of Earthand RE is its radius. Show thatthe net force on a particle ofmass m at a distance x fromthe middle of the tunnel isgiven by Fx = -(GmME/RVx and that the motion of the particle istherefore simple harmonic motion. (b) Show that the period of themotion is independent of the length of the tunnel and is given byT = 27TYRE/g(c) Find its numerical value in minutes.

105 ••• MULTISTEP In this problem, derive the expression forthe average power delivered by a driving force to a driven os-cillator (Figure 14--39).

Weak damping,/highQ

/:;.0).-Heavy damping,10wQ

FIGURE 14-39

Problem 105

(a) Show that the instantaneous power input of the drivingforce is given by P = Fv = - AwFo cos wt sin(wt - 0).

(b) Use the identity sin (°1 - (2) = sin °1 cas 02 - cas °1 sin 02 toshow that the equation in Part (a) can be written asP = AwFo sin 0 cos2 wt sin - AwFo cas 0 cas wt sin wt.

(e) Show that the average value of the second term in your re-sult for Part (b) over one or more periods is zero, and thattherefore Pay = !AwFo sin o. I

(d) From Equation 14--56for tan 0, construct a right triangle inwhich the side opposite the angle 0 is bw and the side adjacentis m(w~ - w2), and use this triangle to show that

. bw bwAsm 0 = --------------- = -- SSMVm2(w~ - w2? + b2w2 Fo

(e) Use your result for Part (d) to eliminate wA from your resultfor Part (e), so that the average power input can be written as

P =.!. F~ sin2 0 = .!.[ bw2F~ ]

ay 2 b 2 m2(w~ - w2? + b2w2

106 ••• MULTISTEP In this problem, you are to use the resultof Problem 105 to derive Equation 14--51.At resonance, the de-nominator of the fraction in brackets in Problem 105(e) is b2w~and Pay has its maximum value. For a sharp resonance, the vari-ation in w in the numerator in this equation can be neglected.Then, the power input will be half its maximum value at the val-ues of w, for which the denominator is 2b2w~.(a) Show that w then satisfies m2(w - wo?(w + wo? = b2w~.(b) Using the approximation w + w2 = 2wo' show that

w - Wo = ~b/2m.(c) Express b in terms of Q.(d) Combine the results of Part (b) and Part (e) to show that there

are two values of w for which the power input is half that atresonance and that they are given by

Wo Wow = w - - and w = w --I 0 2Q 2 0 2Q

Therefore, w2 - WI = ~w = wo/Q, which is equivalent toEquation 14-51.

107 ••• SPREADSHEET The Morse potential, which is often usedto model interatomic forces, can be written in the formU(r) = D(l - r{J(r-r,)?, where r is the distance between the twoatomic nuclei. (a) Using a spreadsheet program or graphingcalculator, make a graph of the Morse potential usingD = 5.00eV,{3 = 0.20nm-1, and ro = 0.750nm. (b) Determine theequilibrium separation and "force constant" for small displace-ments from equilibrium for the Morse potential. (e) Determine anexpression for the oscillation frequency for a homonuclear diatomicmolecule (that is, two of the same atoms), where the atoms eachhave mass m.