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College Physics Student Solutions Manual Chapter 16

118

CHAPTER 16: OSCILLATORY MOTION AND WAVES 16.1 HOOKE’S LAW: STRESS AND STRAIN REVISITED

2. It is weigh-‐in time for the local under-‐85-‐kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s law and is depressed 0.75 cm by its maximum load of 120 kg. (a) What is the spring’s effective spring constant? (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-‐85-‐kg team?

Solution (a) Using the equation m 10750.0 and kg 120 where, 2−×==−= xmkxF gives:

( )( ) N/m 101.57m 100.750

m/s 9.80kg 120 52

2

×=×−

−=−=−=−x

mgxFk

The force constant must be a positive number.

(b) ( )( ) kg 77kg 90.76

m/s 9.80m 0048.0N/m 1057.1

2

5

==−

×−=

−=⇒−=

−=

gkxmkxmg

kxF

Yes, he is eligible to play.

5. When an 80.0-‐kg man stands on a pogo stick, the spring is compressed 0.120 m. (a) What is the force constant of the spring? (b) Will the spring be compressed more when he hops down the road?

Solution (a) Using the equation m 0.120and kg 80 where, −==−= xmkxF gives:

( )( ) N/m 106.53m 0.120

m/s 9.80kg 80.0 32

×=−

−=−=−=xmg

xFk


119

(b) Yes, when the man is at his lowest point in his hopping the spring will be compressed the most.

16.2 PERIOD AND FREQUENCY IN OSCILLATIONS

7. What is the period of Hz0.60 electrical power?

Solution Using the equation T

f 1=

where 1s 0.60Hz 0.60 −==f gives:

ms 7.16 s 1067.1Hz 0.60

11 2 =×=== −

fT

16.3 SIMPLE HARMONIC MOTION: A SPECIAL PERIODIC MOTION

15. A 0.500-‐kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change the period to 2.00 s?

Solution Using the equation

kmT π2= for each mass: ,2;2 2

21

1 kmT

kmT ππ ==

So that the ratio of the periods can be written in terms of their masses:

kg) (0.500s 1.50s 2.00 2

1

2

2

12

2

1

2

1 ⎟⎠

⎞⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛=⇒= mTT

mmm

TT

The final mass minus the initial mass gives the mass that must be added:kg 0.389kg 0.500kg 0.889Δ 12 =−=−= mmm

16.4 THE SIMPLE PENDULUM


120

24. What is the period of a 1.00-‐m-‐long pendulum?

Solution Use the equation

gLT π2= , where m 1.00=L :

s 2.01m/s 9.80

m 1.0022 2 === ππgLT

30. (a) What is the effect on the period of a pendulum if you double its length? (b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?

Solution (a) Use the following equation

gLT π2= , and write the new period in terms of the

old period, where LL 2=ʹ′ :

.22222'2

TgL

gLT =⎟

⎟⎠

⎞⎜⎜⎝

⎛== ππ

.2factor aby increases period The

(b) This time, LL 950.0=ʹ′ , so that

.97.5 0.950 950.02' TTgLT === π

16.5 ENERGY AND THE SIMPLE HARMONIC OSCILLATOR

36. Engineering Application Near the top of the Citigroup Center building in New York City, there is an object with mass of kg1000.4 5× on springs that have adjustable force constants. Its function is to dampen wind-‐driven oscillations of the building by oscillating at the same frequency as the building is being driven—the driving force is transferred to the object, which oscillates instead of the entire building. (a) What effective force constant should the springs have to make the object oscillate with a period of 2.00 s? (b) What energy is stored in the springs for a 2.00-‐m displacement


121

from equilibrium?

Solution (a) Using the equation ,2

kmT π= where s 00.2 and kg 1000.4 6 =×= Tm , gives:

N/m 103.95s) (2.00

kg) 10(4.004π4 62

52

2

2

×=×

==Tmk π

(b) Using the equation

2el21PE kx= , where m 00.2=x , gives:

J 107.90m) N/m)(2.00 10(3.9521

21PE 6262

el ×=×== kx

16.6 UNIFORM CIRCULAR MOTION AND SIMPLE HARMONIC MOTION

37. (a) What is the maximum velocity of an 85.0-‐kg person bouncing on a bathroom scale having a force constant of N/m1050.1 6× , if the amplitude of the bounce is 0.200 cm? (b) What is the maximum energy stored in the spring?

Solution (a) Use the equation X

mkv =max , since the person bounces in simple harmonic

motion, where :m 100.200 2−×=X

m/s 0.266m) 10(0.200kg 85.0

N/m 101.50 26

max =××

== −Xmkv

(b)

( )( ) J 00.3m 10200.0N/m 1050.121

21PE 2262 =××== −kX


122

16.8 FORCED OSCILLATIONS AND RESONANCE

45. Suppose you have a 0.750-‐ kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is simple friction between the object and surface with a static coefficient of friction 100.0s =µ . (a) How far can the spring be

stretched without moving the mass? (b) If the object is set into oscillation with an amplitude twice the distance found in part (a), and the kinetic coefficient of friction is

0850.0k =µ , what total distance does it travel before stopping? Assume it starts at the maximum amplitude.

Solution (a) We know that there are two forces acting on the mass in the horizontal direction: ,0net s =−= kxNF µ where x is the distance the spring is stretched. We know

that N/m, 150 ,kg 0.750 ,100.0s === kmµ and since the mass is on a horizontal

surface, mgN = . We can then solve for x :

m 104.90N/m 150

)m/s kg)(9.80 750(0.100)(0. 32

s −×===kmgx µ

So, the maximum distance the spring can be stretched without moving the mass is .m 104.90 3−×

(b) From Example 16.7,

( ) ( ) m 101.15)m/s kg)(9.80 .750(0.0850)(0

m) 10N/m)(9.80 0.5(15022/1 22

23

k

2−

−

×=×

==mgxkd

µ

16.9 WAVES

50. How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 40.0 m and a propagation speed of 5.00 m/s?

Solution Use the equation λfv =w :


123

Hz 0.125m 40.0

m/s 5.00 ww ===⇒=

λλ

vffv

Now that we know the frequency, we can calculate the number of oscillation:

times50.7s) Hz)(60.0 (0.125 === ftN

55. Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 340 m/s?

Solution Use the definition of velocity, ,tdv =

given the wave velocity and the time:

cm 34.0m 0.340s) 10m/s)(1.00 (340 3w ==×== −tvd

Therefore, objects that are 34.0 cm apart (or farther) will produce sounds that the human ear can distinguish.

16.10 SUPERPOSITION AND INTERFERENCE

62. Three adjacent keys on a piano (F, F-‐sharp, and G) are struck simultaneously, producing frequencies of 349, 370, and 392 Hz. What beat frequencies are produced by this discordant combination?

Solution There will be three different beat frequencies because of the interactions between the three different frequencies.

Using the equation 21B fff −= gives:


124

Hz 43Hz 349Hz 392

Hz 22Hz 370Hz 392

,Hz 21Hz 349Hz 370

3B

2B

1B

=−=

=−=

=−=

f

f

f

16.11 ENERGY IN WAVES: INTENSITY

71. Medical Application (a) What is the intensity in 2W/m of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of energy into a circular spot 2.00 mm in diameter in 4.00 s? (b) Discuss how this intensity compares to the average intensity of sunlight (about 2W/m1 ) and the implications that would have if the laser beam entered your eye. Note how your answer depends on the time duration of the exposure.

Solution (a) Using the equations API = and

tWP = , we see that

( )27

232

2

W/m101.11s) (4.00m 102.000.900π

J 500900.0

: thatso ,900.0)900.0(

×=×

==

===

−trWI

trItIAPtW

π

π

(b) The intensity of a laser is about 410 times that of the sun, so clearly lasers can be very damaging if they enter your eye! This means that starting into a laser for one second is equivalent to starting at the sun for 11 hours without blinking!

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