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CHAPTER 15 OSCILLATIONS

Select odd-numbered solutions, marked with a dagger (), appear in the Student Solutions

Manual, available for purchase. Answers to all solutions below are underscored.

15-1. (a) A = 3.0 m.

angular frequency = 2.0 rad/sec,

frequency2 1

Hz 0.318 Hz2 2

f

= = = =

period T= 1/f= = 3.14 s.

(b) x = 0 when 2.0t= /2, 3/2, 5/2, The first time this occurs is s 0.785 s.4

t

= =

The turning points occur whenx = A. This happens when 2.0t= 0, , 2, 3, The first time

this occurs aftert= 0 is s 1.57 s.2

t

= =

15-2. (a) t= 0,x = 0.6 m.

t= 0.50 s,x =0.5

0.6 m cos 0.6 m cos2 4

= =

i0.42 m.

t= 1.00 s,x = 0.6 m cos2

= 0 m.

(b) 0.6 sin (0.3 ) sin .2 2 2

dx t t v

dt

= = =

at t= 0 sec, v = 0.

at t= 0.50 sec, v =0.50

0.3 sin 0.3 sin 0.67 m/s.2 4

= =

i

at t= 1.00 sec, v = 0.3 sin 0.3 sin 0.94 m/s.2 2 = =

(c)2

0.6 cos 1.48cos4 2 2

dv t t a

dt

= = =

at t= 0 sec, a = 1.48 m/s2

at t= 0.50 sec, 21.48cos 1.05 m/s4

a

= =

at t= 1.00 sec, a = 0.

15-3. (a)1 1

Hz 0.83 Hz1.2

fT

= = =

angular frequency = 2f= rad/s 5.2 rad/s0.6

=(b) A = 0.20 m.

(c) The equation of motion is2

cos 0.20 cos .0.6

x A t tT

= =

Ifx = 0, cos( )

0.6t

= 0

0.6 2t

= t= 0.30 s. Because the period is 1.2 s, the particle will also pass throughx = 0 at

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t= 0.90 s, 1.50 s, 2.10 s, etc. (In SHM the particle passes through the equilibrium position twice

during each cycle.)

Forx = 0.10 m, 0.10 0.2 = cos( )0.6

t

2 4 8 10cos 0.5 , , , ,

0.6 0.6 3 3 3 3

t t

= =

t = 0.40 s, 0.80 s, 1.60 s, 2.00 s, ....

(d) The speed of the motion is: 0.2 sin sin0.6 0.6 3 0.6

dx t t v

dt

= = =

From (c) we know that the first time x = 0 corresponds to t= 0.30 s. At this time,

(0.3)sin m/s 1.05 m/s.

3 0.6 3v

= = =

The first time x = 0.1 m corresponds to t= 0.40 s. At this time,(0.4)

sin 0.91 m/s.3 0.6

v

= =

15-4. A = 4.0 cm.

period T= 1 60sec/ min600rev / minf

= = 0.10 s

frequencyf = 10 Hz

Angular frequency = 2f= 20 rad/s

15-5. The equation of motion of the particle is cos(100 )x t=

(a) The speed of the satellite is: 0.8 100 80 m/s 251 m/ssv r = = = =

(b) The speed of the particle at this point is:

0.0050.8 100 sin(100 ) 80 sin

2t sdx

v tdt

== = = = 80 m/s 251 m/s, =

which is the same as the speed of the satellite.

15-6. x = 1.2 mm cos t; = 2f= 800/s

x = ( )31.2 10 m cos (880 t)

dx

dt= 1.2 103m 880 /s (sin 880 t)

max

dx

dt= ( )31.2 10 880 m/s = 3.3 m/s

2

2

d x

dt= 1.2 103m (880 )2/s cos (880 t)

2

2

max

d x

dt= 1.2 103 (880)2m/s2 = 3 29.2 10 m/s

15-7. frequencyf= 80/min = 11 min

80 min 1.33 Hz.60 s

= = 2 f = 8.38 rad/sec. The magnitude of

the force is2 1 2(6.0 kg)(8.38 s ) (0.50 m) 211 N.max maxF ma m A

= = = =

There is insufficient information given to find a velocity. The maximum speed is18.38 s 0.50 m 4.2 m/smaxv A

= = =

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15-8. cos( ); sin( )dx

x A t v A tdt

= + = = +

At t= 0,x = 0, which meansAcos = 0. This means = /2, 3/2, etc. v(0)= v0 > 0, which

implies sin < 0. Thus the smallest possible values are = /2 or +3/2.

15-9. cos( )x A t = +

When t= 0, 00 cos cos xx x AA

= = =

sin( )dx

v A tdt

= = +

When t= 0, 00 sin sinv

v v AA

= = =

Because 2 2sin cos 1, + = 2 2

0 0 1v x

A A

+ =

22 2 0

0 2

vA x

= +

22 00 2

,v

A x

= +

The phase angle can be expressed three different ways:

1 10 0

22 00 2

cos cosx x

A vx

= =

+

(from the equation forx(0))

1 1 10 0 0

2 2 22 0 0 00 2

sin sin sin( )

v v v

A v x vx

= = = + +

(from the equation forv(0))

1 0

0

tanv

x

=

(by dividing v(0) byx(0))

15-10. 32 2 261.7 1.644 10 rad/sf = = =

period 31 1

3.821 10 s261.7 Hz

Tf

= = =

3 33 10 m 1.644 10 rad/smaxv A= = = 4.93 m/s

2 3 3 23 10 m (1.644 10 rad/s)maxa A= = = 3 28.11 10 m/s

15-11. f= 250 Hz. 2 2 250 rad/s 500 rad/sf = = =2

maxa A= Ifamax =g= 9.81m/s2, the beads start to lift off.

2 22 2 6

2 2

9.81 m/s 9.81 m/s9.81m/s 3.98 10 m.

(500 rad/s)A A

= = = =

15-12. 10 cos cm4

x t

=

10 sin 7.85sin cm/s4 4 4

x

dxv t t

dt

= = =

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22 2

20.617 cm/sx

dv d xa x x

dt dt = = = =

At t= 1.0 s, 10 cos 7.07 cm4

x

= =

7.85sin cm/s 5.55 cm/s4x

v = =

2 2 20.617 cm/s (0.617)(7.07)cm/s 4.36 cm/sxa x= = =

At t= 2.0 s, 10 cos 02

x

= =

7.85sin cm/s 7.85 cm/s2

xv = =

20.617 cm/s 0xa x= =

15-13. cos( ),x A t = + sin( )dx

v A tdt

= = +

at t= 0,x = 0 0 cos = can be /2, 3/2.

at t= 0, v > 0 sin 0A > 3

2

=

15-14. For simple harmonic motion,x =A cos (t+ ) and2

2

d x

dt= 2A cos (t+ ).

Hence, the peak acceleration is 2A, which yields 2A = 0.4 g or

A = 0.4g/2 =( )

2

2

0.4 9.8 m/s1.1 m

2 0.3/s=

Thus, the amplitude of the motion is 1.1 m, i.e., 2.2 m from one extreme of the motion to the

other.

15-15. f= 3.0 Hz, so = 2f= 6.0rad/s.

Letx =A cos (t+ ) andt

dx

d= A sin (t+ ).

t= 0,x =A cos = 0.20 m (i)

dx

dt= A sin = 4.0 m/s (ii)

(ii)/(i) gives tan = 20 rad/s

tan = 20

6= 1.06

= 0.815 radian

From (i)A = 0.20 m/cos = 0.20 m/cos (0.815) = 0.292 m.

= 6.0= 18.8 rad/sTherefore, 0.292 cos(18.8 0.815)x t=

(b) At the turning point

dx

dt= 18.8 (0.292) sin (18.8t 0.815) = 5.48 sin (18.8t 0.815) = 0

Therefore,

18.8t 0.815 = 0 t=0.815 rad

18.8 rad / s= 0.0434 s.

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a =2

2

d x

dt= (18.8 rad/s)2(0.292 m) cos (18.8t 0.815)

= 103 cos [18.8 (0.0432) 0.815]

= 103 cos 0 = 103 m/s2.

15-16. T= 0.70 s2 2

20.70 s

fT

= = = = 8.97 rad/s

2k m= = 70 kg 8.972 (rad/s)2 = 35.63 10 N/m

15-17. BecauseF= kx,2

660 tons 1000 kg/ton 9.81 m/s 2.80 10 N/m0.21 m

Fk

x= = =

angular frequency =6

3

2.8 10 N/m

7.10 10 kg

k

m= = 19.9 rad/s

frequency19.9 rad/s

2 2 radf

= = = 3.16 Hz

15-18. When two springs are connected in parallel, ktotal= k1+ k2.Since the two springs are identical, so: ktotal= 2k

the angular frequency2

2 2total one springk k k

m m m = = = =

2 2 8 Hz 11.3 Hz2

one springf f

= = = =

15-19. =m

kork= 2m, where =

2.

T

Then k=

2

T

2m = 42

2

m

T

Thus, kfor all the four springs together = 42 1100 kg/(0.75 s)2 = 77,200 kg/s2

Because the springs are in parallel, the spring constant for each

= 4 21.9 10 kg/s4k =

15-20. 1 = ,k

m2 =

2

(the are equal because the "springs" are equal)k

km

1

2

= 2

1

,m

m1 =

2

1

m

m2 21

22

1

(2 )m

m f1 =

2

1

m

mf2 .

Letf1, m1be the frequency, mass of D2, andf2, m1 be that of H.

Therefore,

f1 =14 131 (1.31 10 /s) 9.27 10 /s

1.998=

15-21. Because D is two times heavier than H, the ratio of lengthsfrom H to the CM and D to the CM is 2:1. Let be the

angular frequency.

Then = H

H

k

m= D

D

.k

mBy their lengths, kH =

3,

2K

kD

= 3Kwhere kis the spring constant of the entire spring.

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=H

3

2

k

m=

H

3

2

k

m=

H

3

2

k

m

But as shown in the example, each

hydrogen atom feels a spring constant

of 2k. Therefore,

=H

3

2

k

m=

H

3 2

2

k

m=

3

2H

where H is the angular frequency

of hydrogen. So,

f=3

2fH =

3

21.31 1014/s 141.13 10 /s.=

15-22. cos( ),x A t = + sin( )dx

v A tdt

= = +

Suppose at time t= t1, ,2

Ax = v = 25 cm/s.

1

1cos( ) cos( )

2 2

AA t t = + + =

1 1

25 cm/s25 cm/s sin( ) sin( )

AA t t

= + + =

Because 2 21 1sin ( ) cos ( ) 1t t + + + =

therefore, 2 225 cm/s 1

( ) ( ) 1A 2

+ = 225 cm/s 1 3

( ) 1A 4 4

= =

2 2 2 2

2 2 2

4 1 4 1 625cm /s 625cm /s

3 3 15 cmA = = = 1.92 rad/s

2

T

= = 3.27 s

Because ,k

m = so

2 2 2 2

20 N/m

1.92 rad /s

km

= = = 5.43 kg.

15-23. cos( ),x A t = + sin( ),dx

v A tdt

= = + 2 cos( )dv

a A tdt

= = +

Because the mass is released from rest at a displacement of 20 cm from equilibrium, that will be

the amplitude:A =0.20 m.

8.0 N/m7.30

0.15 kg

k

m = = = rad/s

At t= 0, v = 0, so 0 sin 0A = =

Therefore, cos ,x A t= sin ,dx

v A tdt

= = 2 cosdv

a A tdt

= =

The maximumspeedis

0.20 m 7.30 rad/smaxv A= = = 1.46 m/s

The magnitude of the maximum acceleration is2 20.20 m (7.30 rad/s)maxa A= = = 10.7 m/s

2

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15-24.0.2 9.81 N

0.15 m

Fk

x= = = 13.08 N/m

13.08 N/m

8.090.2 kg

k

m = = = rad/s

2 2 rad

8.09 rad/sT

= = = 0.78 s.

15-25. Angular frequency5

4

3

6.0 10 N/m3.64 10

0.5 10 kg

k

m

= = = rad/s

frequencyf=4

33.64 10 rad/s 5.51 10 Hz.2 2 rad

= =

Because1

,2 2

kf

m

= = suppose frequency changes a small amount df, then dfcan be

calculated by taking the differential off:

3 / 24

k

df dmm=

The ratio3 / 24

21

2

kdm

df dmm

mk

m

= =

Therefore, 2 .dm df

m f=

Ifdf

f= 0.01%, then 2 ( 0.01%) 0.02%

dm

m= =

The change of mass: 40.02 0.02% 0.5 g 1.0 10 gm m = = =

Assume the thickness of the deposited film is t, then

t4

6

2 3

1 10 g Area density 6.7 10 cm

Area density 2.0 cm 7.5 g/cm

mm t

= = = =

15-26.3 2 2 10 29 (10 ) m 22 10 N/m

2.0 m

Ayk

L

= = = 9.9 105 N/m

51 1 9.9 10 N/m

2 2 2 15 kg

kf

m

= = = = 40.9 Hz

15-27. Equilibrium atx =F

k=

mg

k=

2.5 9.8 N

90 N/m= 0.27 m

of spring =k

m=

90

2.5= 6/s

Letx be positive upward. Then the

ball is let go atx = 0.27 m at v = 0.

These initial conditions are satisfied by

x =A cos (t+ ) 0.27 cos 6t=

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15-28. (a) =k

m= ( )41.1 10 N/m /14 kg = 28/s. Thus v =

2

= 4.5 Hz.

(b) The wheel has to go round at a frequency 4.5 Hz or = 28/s. Its speed is then

r= (28/s) (0.61/2)m 8.5 m/s.= Oscillations get bigger and bigger.

15-29. Letx be the direction as shown, with zero at the unstretched point on the spring. Then,

Fspring= kx.

Fgravity = mgsin .

At equilibrium, kx = mgsin , orx ( sin ) /mg k=

Fext= kx + mgsin = m2

2

d x

dt

(1)

Let x =x mgsin /k. Therefore,x =x + mgsin /k. Substituting into (1) gives

m2

2

sin'

d mgx

dt k

+

= m

2

2

'd x

dt= kx'

Therefore, the motion is simple harmonic with frequency1

2

k

m

15-30. Letx1,x2 be the coordinates as shown. The force on 1 when displacedx1 is

kx1 + k ( )2 1 .x x The force on 2 is kx2 + k ( )1 2 .x x

The equations of motion are then

m2

12

d xdt

= kx1 + k ( )2 1x x = 2kx1 + kx2 (1)

m2

2

2

d x

dt= kx2 + k ( )1 2x x = kx1 2kx2 (2)

(a) If v1 = v2, by symmetry,x1 = x2. Substituting gives:

m2

1

2

d x

dt= 3kx1 . Therefore,

3,

k

m = or

3 /

2

k mf

=

The positions are given by cos( ),n nx A t = + assuming both masses vibrate with the same

amplitude. n = 1,2 identifies which mass is which. Assume that when m1 is at its equilibrium

position, its moving to the right so v1 > 0. Then 1 2,

2 2

= = + with the angular frequency

given. The two masses are moving 180 out of phase with each other.

(b) If v1 = v2, then both will move together at the same rate to the right x1 =x2. By (1) and (2),

m2

1

2

d x

dt= kx, = ,

k

mf=

/

2

k m

The positions are given by cos( ),n nx A t = + again assuming both masses vibrate with the

same amplitude. Assume that when m1 is at its equilibrium position, its moving to the right so

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v1 > 0. Then 1 22

= = with the angular frequency given. The masses are moving in phase

with each other.

15-31. Force of spring = kx, wherex = 0 at equilibrium.Iof each wheel

=1

2MR2. Net force on cart =Fspring +Ffriction. Let the static

frictional force on each wheel =F. Then =FR andI = =FR.

And for the cart to roll

without slipping, a =2

2

d x

dt=R.

Thus,F=I

R

=

2

2 2R

I d x

dt=

1

2M

2

2

d x

dt

and the frictional force

on all four wheels = 2M2

2.

d x

dt

The net force on the cart

= kx 2M2

2

d x

dt

(minus because friction opposes motion)

Then,Fnet = kx 2M2

2

d x

dt= (m + 4M)

2

2

d x

dt (m + 6M)

2

2

d x

dt= kx

Therefore,

( 6 )

k

m M =

+

/( 6 )

2

k m Mf

+=

15-32. (a) E=1

2kA2 where k= m2 = 42f2m =

2

2

4 m

T

=

2

2 2

4 0.24 kg

1.2 s

= 6.58 N/m

Therefore,E= ( ) 21

6.58 N/m (0.20 m)2

= 0.132 J

(b) Since the mass is at its maximum displacement when t= 0, its kinetic energy is zero

wheneverx = A, ort= 0, T/2, T, 3T/2, . The potential energy is zero wheneverx = 0, ort=

T/4, 3T/4, 5T/4, .

(c) WhenK= U, each type of energy is equal to half the total energy:

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2 2

2 2 4 2

E kx kA AU x= = =

( )13 5

cos cos 1/ 2 , , ,2 2 4 2 4 2 42

A T T T TA t t

= = =

or

t= T/8, 3T/8, 5T/8, 7T/8, .

15-33. E=1

2KA2; k= m

2; 42f2m.

ThenE= 22f2mA2 =

22

2

2 (0.60)

2

(8.0)kg (0.25 m)2

3.6 J=

15-34. Energy 2 21 1

.2 2

E K U mv kx= + = +

From energy conservation, this quantity is constant. So we can use the values ofKand Uatx = 0

(equilibrium position) to find the total energy.

Atx = 0, 2 2 21 1 1

2 kg (3 m) 9 J; 0 9.0 J.2 2 2

K mv U kx E= = = = = =

Whenx = A,K= 0 andE= U, so 23

1 2 2 9.0 J 0.15 m.2 8.0 10 M/m

EE kA Ak

= = = =

15-35. If mass is added at the turning point, amplitude wont change.

Because1

2 2

kf

m

= = ,

ifmnew = 2 mold,1 1 1 1

3.02 2 2 2 2

new old

new old

k kf f

m m = = = = Hz = 2.12 Hz.

Because the amplitude does not change,E= 21

won't change.2

kA

Because ,max

v A= (2 ) 0.15 m 2 2.12 Hzmaxnew new new

v A A f = = = = 2.0 m/s.

Because 2 ,maxa A= 2 2 2(2 ) 0.15 m (2 2.12 Hz)maxnew new newa A A f = = = = 27 m/s

2

15-36. E=1

2kA2 for each, so thatEtotal = kA

2 where kis spring constant for each atom.

E= ( )31.13 10 N/m ( )2

100.5 10 m2 182.8 10 J. No.=

15-37.2 2

4.191.5 secT

= = = rad/sec

k

m= 4.19 rad/sec k= m2 = 0.5 kg (4.19)2 rad2/sec2 = 8.78 N/m

E=2 21 1 N 8.78

2 2 mkA A= i 2 2 0.5 J8.78 N/m

EAk= =

= 0.34 m

15-38. WhenK=1

,2

U x= x0=0.15 m, v = v0=25 cm/s = 0.25 m/s

2 2 2 22 2 2 00 0 0 2 2 2 2

0

1 1 1 1 0.25 m /s 1 2 2 5.55

2 2 2 4 0.15 m s

vkmv kx kx

m x= = = = =

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2

15.55

s

k

m = = = 2.36 rad/s

T=2 2 rad

2.36 rad/s

= = 2.67 s

15-39. After the mass has undergone a displacement din the direction of the force, the work done by the

force is .FW Fd= If there were no friction and the spring were not present, all of this work

would be converted to kinetic energy. But some of the work is stored as potential energy in the

spring: 21

.2

U kd= Thus the kinetic energy is 21

2K Fd kd=

15-40. Kinetic energy of mass when it hits the springK=1

2mv2 = 6.0 J.

Using conservation of energy,

K= Uspring =1

2kx2 =

1(300)

2x2 = 6

x = 0.173 m.

The motion of the mass is simple harmonic with period T= 2 /m k = 0.618 s. The time for themass to stop is T/4 = 0.157 s

15-41. The lengths of spring from the center of mass to m1 and to m2 are in the ratio m2:m1. Since the

center of mass remains fixed, the motion of each mass can be treated as though that mass were

attached to one end of a shortened spring, whose other end is fixed (see figure).

The segment of spring belonging to mass m1 is a fraction

( )2

1 2

m

m m+

of the total spring. The spring

constant of this segment is therefore

k 1 22

m m

m

+

(see Example 6.10). The frequency of oscillation of this

mass is then

=( )

1

spring constant

m=

( )1 2

1 2

k m m

m m

+

15-42. (a) Take a small element dx of spring at a distancex

from the support. That element has mass (dx/e')m',m'=

mass of the spring.

Then the kinetic energy of that mass element is1

2 '

dx

l

m

2

'

xv

l

Since the velocity of that point is'

x

lv

d(KE) =2 2

3

1 ' x

2 '

m v

ldx and

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KEof entire spring =2 2

30 0

1 x(KE)

2

l x l

x

m vd

l

=

=

=

dx =

2 32

3

1 1

2 3 6

m v lm v

l

=

Then the total kinetic energy =1

2mv2 +

1

6m'v2 =

1 1

2 3m m

+

v2

(b) Let angular frequency = . The motion isA cos tfor some amplitudeA.

v2 =A22 sin2t, so that

KE =1

2A2

1

3m m

+

2 sin2t.

KEmax =1

2A2

21

3m m

+

This must equal the maximum potential energy

= 2max1

2k =

1

2kA2.

Therefore,

1

2kA2 =

1

2A22

1

3m m

+ 2 = 13

k

m m +

= 1

3

k

m m +

(c) new

=

1/( )

3

/

k m m

k m

+ =

1

3

m

m m +

=400

5400

3+

= 0.998.

Thus, it is (1 0.998) 100% = 0.2% smaller.

15-43.2

27 m2 2

9.81 m/s

lT

g = = = 10.4 s.

15-44.1 min 60 s

8 rev 8 revT = = = 7.5 s

Because 2 ,l

Tg

= therefore,2 2 2 2

2 2

7.5 s 9.81 m/s

(2 ) (2 )

T gl

= =

i= 14.0 m

15-45. 2l

Tg

= =2

300 m2

9.81 m/s = 34.8 s.

15-46. 2l

Tg

=

Because the length is the same,Jupiter Earth

Earth Jupiter

9.81

0.62924.8

T g

T g= = =

TJupiter= 0.629 TEarth = 0.629 2.00 = 1.26 s.

15-47. T= 2 ,l

gso thatf=

1

T=

1

2

g

l=

21 9.81 m/s0.188 Hz.

2 7 m

=

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15-48. Using T= 2l

g e =

2

24

Tg

Paris l=2 2

2

2.000(9.809)

4

s

m/s2 = 0.9939m

Buenos Aires l=2

9.797

m = 0.9926m

Washington D.C. l=2

9.801

m = 0.9930m

15-49. T1 = 21

l

gT2 = 2

2

l

g

Therefore, 2

1

T

T= 1

2

g

gT2 = period in Austin

T1 = period in N.Y.

2

1

T

T=

9.803

9.793= 1.00051

so that 1 min in Austin is really 1.00051 min. Since 1 day = 24 60 min = 1440 min, clock will

gox min in 1440 min, where 1.00051x = 1440.

x =1440

1.00051= 1439.27 min or the clock will fall behind 0.73 min/day

T1 = 21

l

g= 2.000 s.

l2 =2

2 1

24

g T

=

2

2

9.793(2.000s)

4 m= 0.9922 m.

So that e = 0.9932 0.9922 m = 6.4 10

3

m = 1.0 mm15-50. If the clock runs 1 min late a day, then T1/T2 = 1440/1439 where T1 = period of clock, T2 = correct

period. But T= 2l

gso,

1

2

T

T= 1

2

l

l 2l =

21

1

Tl

T

l2 = l1

2

2

1

,T

T

where l2 = correct length, l1 = actual length

l2 = 0.994 m

21439

1440

= 0.9926 m. This must be shortened by about 1.4 mm.

15-51. 2l

Tg

= 2 2 2

2

2 2

10 s 9.81 m/s

4 4

Tl g

= = = 24.8 m

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15-52. 2l

Tg

=

2 22

Earth AsteroidEarth Asteroid2 2 24 4 4

T TTl g g g

= = =

2 2

Earthasteroid Earth2 2

Asteroid

10 9.8189

T

g gT= = = 0.124 m/s2

15-53. 2 ,I

Tmgd

= where dis the distance between the point of suspension and the center of mass. In

this case, d= 0.9 m.

I= moment of inertia of the painting = 21

2mR

2

2

11 122 2 2 2 m

2 2 9.81 m/s 0.9 m

mR

T Rmgd gd

= = = =i i 3.0 s

15-54. 2 ITmgd

= =2

2

1 m2 2 29.81 m/s

mR R

mgR g = = = 2.0 s

15-55. (a)2

2,

dI I

dt

= =

whereI= moment of inertia of a disc = 21

2MR

2 2

2

2 2 2

1 20 0

2

d dMR

dt dt MR

+ = + =

The solution to this differential equation is cos( ),A t = +

where 2

2

MR

=

(b) The angular velocity is.

sin( ).d

A tdt

= = + The maximum angularspeedis

.

02 2

2 2max A A

MR MR

= = =

Note that is not the same as !

(c) For max = , 0 02 22 2

1 radianMR MR

= =

15-56. Because T=2

= 0.5 s, and2

,

MR

= then

2

2MR

T

=

Therefore,2 2 2 3 2 2

2 2 2 2

4 4 8 10 kg 0.01 m

0.5 s 0.5 s

MR

= = = 1.26 104 kg m2/s2

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15-57. For a pendulum,2

2

d

dt

=

g

L

= 2

(for small values of). Then

= 0 cos

0.5g

L

t. But the motion

here is not truly simple harmonic,

as not all values ofare possible.

From the diagram we see that

at t= 0, 0 = 10, and that the minimum

value ofis 5, or 0 .2

If we solve 0

2

= 0 cos

0.5g

L

t'fort', then

0.5g

L

t'= 60,

ort'= 1.047

0.5

.L

g

The ball will bounce

back in the same time interval.0.5

The period of the motion is therefore 2 2.09 s.L

tg

=

15-58. Angle of arc = = 0.10.99

mm

=A cos t= max cos t

=g

l

Then =A cos ge

tand

d

dt

= A

g

lsin

g

lt.

d

dt

max =Ag

lso that

vmax = ld

dt

=A gl =10

9.8 0.9999

m/s 0.31m/s=

2

2

d

dt

= A

g

l

cosg

ltso that

2

2

d

dt

max

=Ag

l

2

2

d x

dt= l

2

2

d

dt

=Ag= maxg=10

99

9.8 m/s2 20.99 m/s=

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(b) Force exerted = centripetal force exerted on bob + gravity

=2mv

r+ mg= m

2maxv

l g+= (0.40 kg)

2 2 2(0.31) (m /s )

0.99+ 9.8 4.0N=

At the endpoint, the force is simply

the component of the force in

radial direction= mgcos

= 0.40(9.8) cos10

99

3.9N=

15-59. (a) E=1

2mglA2

A = max = 40 =4

180=

45

E=1

2mgl

2

45

=

1

4050mgl2

= ( )21

(0.120 kg) 9.81 m/s (0.44 m)4050

231.26 10 J=

(b) 2max1

2mv =E, vmax = speed at lowest point

vmax =2E

m=

32 1.26 100.145 m/s

0.120=

15-60. = .mgl

ITherefore, T=

2

= 2

I

mgl

TNBS = 2NBS

;

I

mlg TUSCG = 2

USGC

I

mg

Therefore NBS

USCG

T

T= USCG

NBS

g

ggUSCG =gNBS

2

NBS

USCG

T

T

gUSCG = 9.80095 m/s2

2

2.10356 s

2.10354 s

29.80104 m/s=

Percent change =9.80104 9.80095

9.80095

100% 0.0009%=

15-61. 2I

Tmgd

=

Because the stick swings around one of its ends, 21

.3

I ml= Therefore,

21

232 .3

mll

Tmgd gd

= =

The distance between the point of suspension and the center of mass is d= 0.5 m.

The length of the stick is l= 1 m.

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Then2

2 2 1 m

3 3 9.81 m/s 0.5 m

lT

gd

= = = 1.64 s

15-62. Iof cylinder about CM through axis perpendicular to

longitudinal is

m

2 2

4 12

a l

+ where a = radius, l= length.

Therefore,Iabout pivot

=Irod +Icylinder

= 2 21 1

3 4md mb

+

+M2 2

4 2

a l

I

+

+M

2

ld

+

2

where m = mass of rod,

M= mass of cylinder.

Center of mass of pendulum is

x =

( / 2) ( / 2)m d M d l

m M

+ +

+m = b

2d,M= a2e

where = density of brass.

Therefore,x =

( )

( )

2 2

2 2

2 2

d lb d a l d

b d a l

+ +

+

=

2 2 2 22

2 2

2 2

b d a l a ld

b d a l

+ +

+

I= b2d2 21 1

3 4d b

+

+ a2l

2 22

4 3

a ld dl

+ + +

( )gx

m MI

+ =

2 2 22 2

2 22 2 2 2 2

2 2

1 1

3 4 4 3

b d la a ld g

a lb d d b a l d dl

+ +

+ + + +

d= 90.00 cm, e = 20.00 cm, a = 3.00 cm, b = 0.50 cm andg= 980.7 cm/s2

=( )

( )

7

4 6

1.865 10

6.075 10 1.806 10 rad/s+= 3.161 rad/s

T=2

= 1.988s

15-63. Tbrass = 2brass

l

gTiron = 2

iron

l

g

T= 21.800

9.81= 2.691 s. Then the number of cycles after 12 min =

( )12 60

2691= 267.5 cycles.

1/4 of a one-way swing is T/8 = 0.13 cycle. Since the pendulums are 1/4 swing out of step, the

ratio of one period to the other cannot be different by more than (267.5/267.63).

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1

2

T

T=

267.5

267.63= 2

1

.g

gTherefore

g2 =g1267.5

267.63

2g2 = 0.999g1

Largest difference is 0.001g=

3 2

9.8 10 m/s

15-64. Moment of inertia

I=1

3M1l

2 +1

12M2d

2 +M2

2

2

dl

+

= ( )21

6.8 (0.43)3

+1

(4.1)(0.46)12

2

+ 4.1 (0.66)2 = 2.3 kg m2

Center of mass is at

1 2

1 2

1

2 2

dM M l

M M

+ +

+

= 0.38 m

=( )1 2 (0.38)(9.8)M M

I

+

=( )10.9 (0.38)(9.8)

2.3= 4.2 rad/s.

Time period T=2

1.5s

= which is approximately the

experimental value.

15-65. =

mgl

I . Approximate the ruler by a thin rod. ThenIthrough

axis (by parallel axis theorem) =1

12md2 + me2 (e,dare as shown)

=2 21

12

mgl

m d l +

=2 21

12

ge

d l +

=2

2 2

9.8m/s 0.2m

1(1.0) 0.2

12

+

= 4.0/s

Therefore T= period = 2 1.6s =

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15-66. (a) =mgl

I

I= (by parallel axis theorem)

=1

2mR2 + ml2

Then =2 21

2

mgl

mR ml +

=2 21

2

gl

R l +

(b)d

de

=

1/ 2

2 2

1

12

2

gl

R l

+

2 2 2

2

2 2

12

2

1

2

g R l ge

R l

+

+

=g2

2 2 2 2

2

2 2

1 1

2 2

1

2

R l gR gl

glR l

+

+

This equals zero when1

2R2 = l2 or

2

Rl=

15-67. The frequency, =Mgh

I=

2

T

whereM= m1 + m2, and h as shown in the diagram, is the

distance from the center of the rod (the axis of rotation), to the center of mass of the system.Iis

the moment of inertia of the system about the axis of rotation.I= m1L2 + m2L

2 But

( )1 2m m+ h = m1L m2L

h =( )

( )1 2

1 2

.m m L

m m

+

Therefore T= 2I

Mgh

( )1 2

1 2

( )2

m m LT

m m g

+=

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15-68. The figure shows a free-body diagram for the rod. The

torque of the weight about the suspension point is 1/2 mgL

sin , and the moment of inertia is 1/2 ml2. Hence the

equation of rotational motion is

1

3

mL2 =1

2

mgL sin

=3

2

g

Lsin

The acceleration of the center of mass is a = L/2, in the transverse direction (see figure: since

= 0, there is no centripetal acceleration at the initial instant). The equation for translational

motion in this transverse direction is then2

m L= mgsin Fsin . The equation for

translational motion in the direction along the rod is 0 = mgcos Fcos .From these equations, we find

Fsin = mgsin 3

4mgsin =

1

4mgsin (1)

and

Fcos = mgcos (2)

Squaring and adding these, we obtain

F2 =1

( sin )16

mg 2 + (mgcos )2 = (mg)2 215

116

sin

and

F= mg 215116

sin = mg2151 sin 20

16 = 0.94 mg

From the ratio of Equations (1) and (2),

we obtain tan = 1

4tan =

1

4tan 20

= 5.2

Thus the forceFhas a magnitude of 0.96 mg and a direction of 25.2 with the vertical.

15-69. When the door is open, = 54 = = 30 and = 1.8 radian.

Therefore, for the shut door, = 1.8 2

= 0.23 rad. The change in the potential energy between

the open and shut door

U=1

2(0.23)2

1

2(1.8)2 = 2 2

1 (30) (0.23) (1.8)2

= 47.8 J.

From conservation of energy,

K=1

2I2 = U= 47.8 J

=(47.8)(2)

I

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For the doorI=1

3Ml2 =

1(27)(0.91)

3

2 = 7.45 kg m2

Therefore =(47.8)(2)

7.45= 3.6 rad/s

The linear speed of the edge is v = l= 3.6 0.91 3.3m/s=

15-70. According to the text just before Checkup 15.4, the increase in the period of a 30 swing 1.7%.

T= 2 /l g = 2 1.5 / 9.81 = 2.457 s. But that with 30 swing has T= 2.499 s. So after 10 min,

the 5 has swung 600 s/2.457 s = 244.2 times, while that with a 30 amplitude has swung

600 s/2.499 s = 240.0 times. Thus the two are out of synch by about 4 cycles out of 240, which

would be easy to detect with careful counting. However, it is unlikely that Galileo could keep the

amplitude fixed at 30. Damping would reduce the amplitude, causing the period to decrease to its

small amplitude value. Galileo would correctly interpret his result by concluding that the period is

independent of the initial amplitude.

15-71. The restoring force exerted by the rod,F= k.

The restoring torque is therefore kd, where dis the length of the rod. With a mass m attached to

the upper end, the torque exerted by the mass is mgdsin , and this torque opposes the restoringtorque.

The net torque is therefore = mgdsin kd. For small , sin so that d(k mg)

But from Newtons Second Law, =I =I2

2,

d

dt

whereI= md2 for the system, since we are

treating the rod as massless.

The equation of motion for the rod is therefore (for small displacements)2( )md = d(kmg)

+k g

md d

= 0,

which is the equation for simple harmonic motion with 2 = g

d+ k

mdor

=

0.5g k

d md

+

For stable solutions,k

gm

+

must be greater than 0. Therefore the rod becomes unstable when

k

mg= 0; that is, when .

km

g=

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15-72. (a) If the train is at a distancex from the center of the

tunnel (see Figure), then the acceleration of gravity is

3

GM

R

rand the component of this acceleration in the

horizontal direction is

gx = 3

GMrR

cos = 3

GM xrR r

= 3

GM

Rx

(b) The horizontal force on the train is

mgx = 3GMm

Rx

This is a linear restoring force, as in the case of a mass on a spring. The effective spring

constant is k=3

GMm

R. Hence the period of the motion is

T= 2m

k = 2 3/

m

GMm R = 2

3R

GM

(c) The time for a trip from San Francisco to Washington is one-half period,

t=2

T=

3R

GM=

6 3

11 2 2 24

(6.4 10 m)

6.67 10 N m /kg 5.98 10 kg i

= 2.5 103 s = 42 min

(d) For a mass on a spring, the potential energy is1

2kx2, and hence the kinetic energy acquired

when moving fromx =x0 tox = 0 is

2

max

1

2mv =

1

2

2

0kx

from which vmax =2

0kx

m= 0

/

x

m k= 0

2 x

T

where T = 2m

kis the period of the motion.

From Example 1.4, we know that the length of the tunnel is 3.8 103 km, i.e.,x0 =3(3.8 10 km)

.2

From part (b) we know that T= 2 42 min. Hence,

vmax =6

3 32 (3.8 10 m) / 2 2.4 10 m/s 8.5 10 km/h2 42 min

= =

15-73. According to Problem 12.59, the moments of inertia of the cone about an axis through its apex is3/5ML2. To find the period, we need the distance between the apex and the center of mass.

If the half-angle at the apex is , the radius of the cone at a distancezbelow the apex (see

Figure) isR zand the mass in a disk-like segment of radiusR and thickness dzisR2dz.

Hence thez-coordinate of the center of mass is

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40 0

2 3

cm 0 0 32 2

34

43

L L

L L

Lz R dz z dz

z LLR dz z dz

= = = =

Equation (48) for frequency of a physical pendulum tells us

that

= cmMg z

I

=

2

(3 / 4 )

(3 / 5 )

Mg L

ML=

5

4

g

L

The period is then2 4

25

LT

g

= =

15-74. (a) The force at the midpoint is given by:

F(x) = 2( )

GMm

r x

+

+2( )

GMm

r x

dFdx

= ddx

GMm2 2

1 1( ) ( )r x r x

+

= GMm3 3

2 2

( ) ( )r x r x

+ +

By Taylors expansion,

F(x) F(0) + 0xdF

dx=

x = 0 +3

4GMm

rx.

Thus, the force is 4 GMm|x|/r3 in the direction of motion of

the particle.

(b) By symmetry, the force along the line joining the

centers of the two spherical bodies is zero. In a directionperpendicular to this line, it is:

F=2 2( )

GMm

r r

+sin +

2 2( )

GMm

r r

+sin

=2 2 2 2 1/ 2

2

( ) ( )

GMm x

r r r r

+ += 2GMm

2 2 3 / 2( )

x

r r

+

dF

dx= 2GMm

2 2 3 / 2 2 2 1/ 2

2 2 3

3( ) ( ) 2

2 ,( )

r x x r x x

r x

+ +

+

dFdx

|x = 0 = 2GMm3

6rr

=3

2GMmr

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15-75. (a) The tension in the string obeys the equation.

Tmgcos = m 2l= m2l

T= mgcos + m2l

= Ag

lsin

gt

l

Then

T= mgcos cosg

A tl

+ mgA2 sin2g

tl

Using cos = 1 2

2

+ . . . . = 1

1

2A2 cos2

gt

l

= . . . .

we have

T= mg2

2 2 21 cos sin2

A g gt A t

l l

+

Using cos

2

x = 1 sin

2

x, we have

T= mg2

2 2 21 1sin sin2

A g gt A t

l l

+

T= mg2

2

mgA+

3

2mgA2 sin2

gt

l

22 231 sin

2 2

A gT mg A t

l

= +

(b)The sin2 term has a maximum wheng

et= ,

2

so that

.2

lt

g

=

The tension is then

Tmax = mg2

3 231 (1 )2 2

AA mg A

+ = +

15-76. Normal marching pace is around 90 steps per minute, sof 1.5 Hz.

15-77. As in Example 10, well use2

,E

QE

=

where Eis the energy lost during

the first cycle andEis the pendulums initial energy. To find this quantity,

use the hint that E/Eis the same for every cycle. If the energy decays byEto a valueE1 during the first cycle, then it will decay by E1 toE2during the second cycle with E1/E1 = E/E. So after one cycle the energy

would be 1 (1 / )E E E E= and after two cycles the energy would be2

2 1 1 1(1 / ) (1 / ) .E E E E E E E= = Aftern cycles the energy would be

(1 / ) ,nnE E E E= from which we get1/

1 .

n

nEE

E E

=

L

L

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When a pendulum of lengthL is raised through some angle , the mass at the end is lifted through

a heightL L cos . The potential energy of the mass is mgh = mgL(1 cos ). If the pendulum is

released from this height, the equation gives the total energy of the pendulum as it swings.

According to the Math Help box following Eq. (15.46), the cosine function can be approximated

by2

1

2

for small angles given in radians. Using this in the equation for the energy gives

2

.2

mgLE

= For a 1.5-m-long pendulum with an amplitude of 10 (0.175 radian), the initial

energy is2 2(9.81 m/s )(1.5 m)(0.175)

2

mE= = 0.224m J. (No mass is given in the problem, but it

will cancel in the calculation ofQ.) When the amplitude is 4 (0.0698 radian), the final energy is

0.036 J.finalE m= The period of the pendulum is 2 2.46 s,L

Tg

= = so 12 min represents

12 60293 cycles.

2.46 The energy loss per cycle is given by

1/ 293

30.0361 6.22 10 .0.224

E mE m

= =

The final result is 33

2 1.0 10 .6.22 10

Q = =

15-78. (a)2

.E

QE

=

Since 2 ,E A decreasingA by a factor of 2 decreasesEby a factor of 4 toE/4.

The period on Earth for a pendulum with a length of 0.994 m turns out to be 2.00 s (see Example

8), so 13 minutes represents 390 cycles. The energy loss per cycle

is

1/ 3901

1 0.00355,4

E

E

= =

and 32

1.77 10 .0.00355

Q

= =

(b) For an amplitude of 8 = 0.140 radian,2 2 2(1.2 kg)(9.81 m/s )(0.994 m)(0.140)

0.114 J.2 2

maxmgLE

= = =

The energy loss per cycle is2 2

.EE E

E PQ T QT

= = = Thus

4

3

2 (0.114 J)2.02 10 W

(1.77 10 (2.00 s)P

= =

15-79. (a) Follow the method outlined in Problem 77. Here a mass of 25 kg is given. The initial and

final amplitudes are 12 (0.209 radian) and 10 (0.175 radian), respectively.2 2 2(25 kg)(9.81 m/s )(3 m)(0.209)

16.1 J2 2

initmgLE

= = = .

2 2 2(25 kg)(9.81 m/s )(3 m)(0.175)

11.3 J2 2

final

final

mgL

E

= = = after five swings.1/ 5

11.31 0.0684.

16.1

E

E

= =

2 291.9

0.0684

EQ

E

= = =

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(b) The period is2

3 m2 2 3.47 s.

9.81 m/s

LT

g = = = The energy loss per cycle is

2 EE

Q

= from which we get the rate of energy loss:

2 2 (16.1 J) 0.317 W.(91.9)(3.47 s)

EE

T QT

= = =

15-80.2 2

.2 2

initialkA kAE= = 2

.2

inal

final

kAE =

( )

1/101/10 21/ 5

21 1 1 0.95 0.0102.

final finalE AE

E E A

= = = = 2 2

6160.0102

EQ

E

= = =

15-81. 0

0

F AkA Q Q

k F= = =

0.4 m 15 N/m

0.2 N= 30

15-82.18

6 90

3

1 10 N5 10 1 10 m

5 10 N/m

FA Q

k

= = =i

15-83. 21

(2 )2 2

k kk f m

m m

= = = =

f = 100 GHz = 10 109 Hz, m = 1.0 1018 g = 1.0 1021 kgso 9 2 21(2 100 10 Hz) 1.0 10 395k kg = =i N/m

10

10

0

1.0 10 m 395 N/m

1.0 10 N

AkQ

F

= = = 395

15-84. (a) 1

0

0.5,A

A

= so one half of the amplitude is lost in its first oscillation.

(b) SinceE= 21

,2

kA 2

21 1

2

0 0

0.5 0.25E A

E A= = =

Therefore, 75% of the energy is lost in the first oscillation.

(c)1

2 20.75

EQ

E = = =

8.37

15-85. 2E (see problems 77 and 79). Cutting the angular amplitude in half means the energydecreases by a factor of 4, so the final energy isE/4. This loss of energy occurs in about four

cycles, so

1/ 41

1 0.293.4

E

E

= =

Then2

2 21.0.293

EQ

E

= = =

15-86. (a) cos( )x A t = +Since at t= 0, max ,x x= so we should let = 0.

Since T= 0.8 s, so2 2

rad/s0.8T

= =

2

0.2 cos( )0.8

x t

=

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CHAPTER 15

341

(b) At t= 0.1 s,2

0.2cos( 0.1) 0.2 cos( )0.8 4

x

= = = 0.14 m

At t= 0.2 s,2

0.2 cos( 0.2) 0.2 cos( )0.8 2

x

= = = 0 m

At t= 0.3 s,2 3

0.2 cos( 0.3) 0.2 cos( )0.8 4

x

= = = 0.14 m

At t= 0.4 s,2

0.2 cos( 0.4) 0.2cos( )0.8

x

= = = 0.20 m

15-87. A = 1.5 cm

f= 4000 rev/min =4000 rev/ min

60 s/min= 66.7 Hz.

15-88. vmax =A = 1.0 105 m 2 rad/rev 4000 rev/s = 0.25 m/s

15-89. (a) Since the first particle moves according to the equation 0.30cos( ),4

x t

= it reaches its

midpoint whenx = 0 3 5

cos( ) 0 , , ,

4 4 2 2 2

t t

= = t= 2 s, 6 s, 10s, ... . At the

turning points, 0 sin( ) 0 0, , 2 ,4 4

dxt t

dt

= = = t= 0, 4s, 8s, ....

(b) The second particle moves according to the equation ' 0.30sin( ).4

x t

= At its midpoint,

' 0 sin( ) 0 0, , 2 ,4 4x t t

= = =

t= 0, 4 s, 8s, .... At its turning point,' 3 5

0 cos( ) 0 , , ,4 4 2 2 2

dxt t

dt

= = = t= 2 s, 6 s, 10s, ... .

(c) Find the time whenx = x:

0.3cos( ) 0.3sin( ) 0.3cos( ) 0.3cos[ ( 2)]4 4 4 2 4t t t t

= = = We can see that when t = t+2, we will havex = x, i.e., if the first particle passes a certain pointat time t, the second particle passes the same point at time t+2, or 2 s later.

Particle 1Satellite

Midpoint Turning Point

Particle 1

Turning Point

SatelliteParticle 1

Midpoint Turning Point

Satellite

Turning Point

Particle 1Satellite

Midpoint Turning Point

Particle 1

Turning Point

SatelliteParticle 1

Midpoint Turning Point

Satellite

Turning Point

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CHAPTER 15

342

15-90. (a)3.0

2 2f

= = = 0.48 Hz

2 26 3k

k mm

= = = = 54 N/m

vmax =A = 0.2 3 = 0.60 m/s

(b)54 N/m

2.0 kg

k

m = = = 5.20 rad/s.

5.200.83 Hz

2 2f

= = =

vmax =A = 0.2 m 5.2 rad/s = 1.04 m/s

15-91. 6000 rev/min = 100 rev/s, so that v = 100 Hz. Then = wv = 200 /s

Therefore,x =8.50

2cos (200 t) cm (set = 0 since it is arbitrary).

dx

dt= 4.25 cm (200 )/s sin (200 t).

Max at |sin 200 t| = 1 max

2670 cm/s 26.7m/sdx

dt= =

2

2

d x

dt= 2 24.25 cm(200 ) /s cos (200 t).

Max at |cos 200 t| = 1,2

2

max

d x

dt= 1.68 106 cm/s 41.68 10 m/s=

Forcemax = mass accelerationmax = 1.2 kg 1.68 104 m/s2

42.0 10 N=

15-92.2

2 4 .totalm

Tk

= For the 1.00 kg mass,

2

2

4 (1.40 kg)47.4 N/m.

(1.08 s)k

= = When the unknown mass

is used,2 2

2 2

(47.4 N/m)(1.78 s)3.80 kg.

4 4total

kTm

= = = Subtracting the mass of the tray and

straps gives m = 3.40 kg.

15-93. knew = 2knew =new 2 2 old

k k

m m= =

2 2 1.5new old f f= = Hz = 2.12 Hz

15-94. (a) 2k

k mm

= =

Sincef = 2.4 Hz, So 2 4.8f = = rad/s

k = 55 kg (4.8)2 rad2/s2 = 1.25104 N/m

(b)41.25 10 N/m

75 kgnew

k

m = = = 12.9 rad/s

12.9 rad/s

2 2f

= = = 2.05 Hz

15-95.31 1 4.9 10 N/m

2 2 80 kg

kf

m = = = 1.25 Hz

Note that gravity does not affect the frequency of the motion. It only affects the equilibrium point.

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CHAPTER 15

343

15-96. (a) Total energy = 21

2kA

If potential energy Uequals kinetic energyK, then PE =1

2(total energy) =

1

4kA2

21

2kx = 2

1

4kA x2 = 2

1

2A

2

Ax =

i.e., when ,2

Ax = kinetic energy equals potential energy.

(b) if 1

,2

x A= then U=2 2

21 1

2 2 4 8

A kAkx k= =

SinceE= 21

,2

kA U=1

,4

E i.e. 1/4 of total energy is potential energy, 3/4 of total energy is

kinetic energy.

15-97. 2 2 2 21 1 N

6.0 10 0.25 m2 2 m

E kA= = = 18.75 J.

At the equilibrium position, U= 0, soKmax = E= 18.75 J 21 18.75

2maxmv = J

2 2 18.75 J

3.0 kgmax

Ev

m= = = 3.54 m/s

15-98. A = 0.24 m,A will double.

=k

m wont change.

Frequency wont change.

21 42

new new old E kA E= = Energy will quadruple.

vmax = Avmax will double.amax = A

2amax will double.

15-99.2

1.5 m2 2

9.8 m/s

lT

g = = = 2.46s. To change to half the period,

' ' 12 '

2 2 4

T l l l l l

g g g= = = = 0.375 m.

15-100. Before it hits the peg, 11 22 2 2 2 9.8 m/s

l lT

g = = = 4.01 l

22 2

(1/ 4) m2 2 2 2

9.8 m/s

l lT

g

= = = 2.0 l

period = 1 2 6.01T T l+ =

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CHAPTER 15

15-101. When the angle is small,( ) sin ( ) ,Mg R L Mg R L = + +

The moment of inertia is 2 22

( )5

I MR M R L= + +

Therefore, from ,I = we get:2

2 2

2

2( )5

dMR ML

dt

+ ( )Mg R L = +

2

22 2

( )0

2( )

5

d g R L

dtR R L

++ =

+ +

The solution for this differential equation is: cos( ),A t = +

where2 2

( ).

2( )

5

g R L

R R L

+

=+ +

Therefore,

2 22 ( )2 52 .

( )

R R L

T

g R L

+ += =

+15-102. sinmgx =

If the angle is small, mgx

2 21

12I mL mx= +

FromI = we get: 2 21

( ) 012

mL mx gx + + =

2

22 2

01

12

d gx

dtmL mx

+ =

+

2 21

12

gx

mL mx

=+

2 212 122 .

mL mx

Tgx

+= =

To get the minimum period, we find thex that makes 0dT

dx=

d

dx

2 21

12 0

L x

x

+=

12

Lx =

15-103.21 1 1 9.8 m/s

2 0.35 Hz2 2 2 m

l gT f

g T l

= = = = =