amplitude. b) The glider will return to its original position... period is 1.60 s. c) The frequency is the reciprocal... 13.1

13.1: a)
T 
 1 . 14  10
1
4 ( 220 Hz)
b)
 4 . 55  10
1
f
3
3
s, ω 
2π
T
 2 πf  1 . 38  10 rad s.
3
s, ω  2 πf  5 . 53  10 rad s.
3
13.2: a) Since the glider is released form rest, its initial displacement (0.120 m) is the
amplitude. b) The glider will return to its original position after another 0.80 s, so the
period is 1.60 s. c) The frequency is the reciprocal of the period (Eq. (13.2)),
f 
1
1 . 60 s
 0 . 625 Hz.
13.3: The period is
ω
2π
T
 1 . 14  10
0 . 50 s
440
3
s
and the angular frequency is
 5 . 53  10 rad s.
3
13.4: (a) From the graph of its motion, the object completes one full cycle in 2.0 s; its
period is thus 2.0 s and its frequency  1 period  0 . 5 s  1 . (b) The displacement varies
from  0 . 20 m to  0 . 20 m, so the amplitude is 0.20 m. (c) 2.0 s (see part a)
13.5: This displacement is
1
4
of a period.
T  1 f  0 . 200 s, so t  0 . 0500 s.
13.6: The period will be twice the time given as being between the times at which the
glider is at the equilibrium position (see Fig. (13.8));
2
2


2π
 2π 
 ( 0 . 200 kg)  0 . 292 N m.
k ω m 
 m  
 T 
 2 ( 2 . 60 s) 
2
13.7: a) T 
1
f
 0 . 167 s. b) ω  2 πf  37 . 7 rad s. c) m 
k
ω
2
 0 . 084 kg.
13.8: Solving Eq. (13.12) for k,
2
2
 2π 
 2π 
3
k  m
  1 . 05  10 N m.
  ( 0 . 600 kg) 
s
150
.
0
T




13.9: From Eq. (13.12) and Eq. (13.10), T  2 π
0 . 500 kg
140 N m
 0 . 375 s, f 
1
T
 2 . 66 Hz,
ω  2 πf  16 . 7 rad s.
13.10: a) a x 
2
d x
dt
2
  ω A sin( ωt  β )   ω x , so x ( t )
2
ω 
2
k
m
. b) a  2 A ω
ax 
dv x
 ( iω ) Ae
dt
2
2
is a solution to Eq. (13.4) if
a constant, so Eq. (13.4) is not satisfied. c) v x 
i ( ωt  β )
  ω x , so x ( t ) is a solution t
2
dx
dt
 iω
i ( ωt  β )
o Eq. (13.4) if ω  k m 
2
,
13.11: a) x  ( 3 .0 mm) cos ((2 π )(440 Hz) t ) b) (3.0  10  3 m)(2 π )(440 Hz)  8.29 m s,
2
( 3 . 0 mm)(2 π ) ( 440 Hz)
2
 2 . 29  10 m s . c) j ( t )  ( 6 . 34  10 m s ) sin((2 π )(440 Hz) t ),
4
2
7
3
j max  6 . 34  10 m s .
7
3
13.12: a) From Eq. (13.19), A 
v0
ω

v0
k m
 0 . 98 m. b) Equation (13.18) is
indeterminant, but from Eq. (13.14),    2 , and from Eq. (13.17), sin   0 , so    π2 .
c) cos ( ωt  ( π 2 ))   sin ωt , so x  (  0 . 98 m) sin((12.2 rad s ) t )).
13.13: With the same value for ω , Eq. (13.19) gives A  0 . 383 m and Eq. (13.18) gives
and x  (0.383 m) cos (12.2 rad/s) t  1 . 02 rad .

(  4.00 m/s)
  arctan  


(0.200 m)
300 N/m/2.00

  1 . 02 rad  58.5  ,
kg 
and x = (0.383 m) cos ((12.2 rad/s)t + 1.02 rad).
2
13.14: For SHM, a x   ω 2 x   ( 2 πf ) 2 x    2 π ( 2 . 5 Hz)  (1 . 1  10  2 m)   2 . 71 m/s 2 .
b) From Eq. (13.19) the amplitude is 1.46 cm, and from Eq. (13.18) the phase angle is
0.715 rad. The angular frequency is 2 πf  15 . 7 rad/s, so
x  (1 . 46 cm) cos ((15.7 rad/s) t  0 . 715 rad)
v x  (  22 . 9 cm s ) sin ((15.7 rad/s) t  0 . 715 rad)
a x  (  359 cm/s
2
) cos ((15.7 rad/s) t  0 . 715 rad) .
13.15: The equation describing the motion is x  A sin ωt ; this is best found from either
inspection or from Eq. (13.14) (Eq. (13.18) involves an infinite argument of the
arctangent). Even so, x is determined only up to the sign, but that does not affect the
result of this exercise. The distance from the equilibrium position is
A sin  2 π t T
  0 . 600
m  sin  4 π 5   0 . 353 m.
13.16: Empty chair: T  2 π
m
k
2
2
4π m
k 
T
4 π ( 42 . 5 kg)

2
(1.30 s)
 993 N/m
2
With person in chair:
T  2π m k
2
m 
T k
4π
2
2

( 2 . 54 s) ( 993 N/m)
4π
2
 162 kg
m person  162 kg  42 . 5 kg  120 kg
13.17:
T  2π
m k , m  0 . 400 kg
Use a x   2 . 70 m/s
2
to calculate
 kx  ma x gives k  
ma x
k:
( 0 . 400 kg)(  2 . 70 m/s )
2

x
  3 . 60 N/m
0 . 300 m
T  2 π m k  2 . 09 s
13.18: We have v x ( t )  ( 3 . 60 cm/s)sin(( 4.71 s  1 ) t  π 2 ). Comparing this to the general
form of the velocity for SHM:
 ωA  3 . 60 cm/s
ω  4.71 s
1
  π 2
T  2 π ω  2 π 4 . 71 s
(a)
(b)
(c )
A
3 . 60 cm s
ω

3 . 60 cm s
4 . 71 s
1 2
1
1
 1 . 33 s
 0 . 764 cm
a max  ω A  ( 4 . 71 s ) ( 0 . 764 cm )  16 . 9 cm s
2
2
13.19: a) x ( t )  ( 7 . 40 cm) cos((4.16 rad s) t  2 . 42 rad)
When t  T , (4.16 rad s) T  2 π so T  1.51 s
b) T  2 π
m k so k  m ( 2 π T )  26 . 0 N m
2
c) A  7 . 40 cm  0 . 0740 m
1
2
mv 
2
1
2
kx 
2
1
2
kA gives v max  A k m  0 . 308 m s
2
d) F   kx so F max  kA  1 . 92 N
at t  1 . 00 s gives x   0 . 0125 m
e) x ( t ) evaluated
v  k m
A  x   26 . 0 1 . 50
2
( 0 . 0740 )  ( 0 . 0125 ) m s   0 . 303 m s
2
2
2
Speed is 0.303 m s .
a   kx m   ( 26 . 0 1 . 50 )(  0 . 0125 ) m s   0 . 216 m s
2
2
13.20: See Exercise 13.15;
t  ( arccos(  1.5 6))( 0.3 (2 π ))  0.0871 s.
13.21: a) Dividing Eq. (13.17) by ω ,
v0
x 0  A cos  ,
ω
  A sin .
Squaring and adding,
2
x 
2
0
v0
ω
 A ,
2
2
which is the same as Eq. (13.19). b) At time t  0 , Eq. (13.21) becomes
1
kA 
2
2
1
2
mv 0 
2
1
2
kx 0 
2
1 k
2ω
v0 
2
2
1
2
2
kx 0 ,
where m  kω (Eq. (13.10)) has been used. Dividing by k 2 gives Eq. (13.19).
2
13.22: a) v max  ( 2 πf ) A  ( 2 π ( 392 Hz))(0.60  10  3 m)  1.48 m s.
b) K max 
13.23: a) Setting
1
2
1
2
m (V max ) 
2
mv

2
1
2
kx
2
1
( 2 . 7  10
5
kg)(1.48
m s)
2
 2 . 96  10
5
J.
2
in Eq. (13.21) and solving for x gives x  
A
2
.
Eliminating x in favor of v with the same relation gives v x   kA 2 2 m   ωA2 . b) This
happens four times each cycle, corresponding the four possible combinations of + and –
in the results of part (a). The time between the occurrences is one-fourth of a period or
T /4 
2π
4ω

π
2ω
. c) U 
1
4
E, K 
3
4
E
U

kA
8
2
,K 
3 kA
8
2

13.24: a) From Eq. (13.23),
k
v max 
450 N m
A
m
( 0 . 040 m)  1.20 m/s.
0.500 kg
b) From Eq. (13.22),
450 N
v 
2
( 0 . 040 m)
 (  0 . 015 m)
 1 . 11 m/s.
2
0.500 kg
c) The extremes of acceleration occur at the extremes of motion, when x   A , and
a max 
kA

( 450 N/m)(0.040
m
d) From Eq. (13.4), a x  
m)
 36 m/s
2
(0.500 kg)
( 450 N/m)(  0.015 m)
 13 . 5 m/s .
2
(0.500 kg)
e) From Eq. (13.31), E  12 ( 450 N/m)(0.040
m)
2
 0 . 36 J.
13.25: a) a max  ω 2 A  ( 2 f ) 2 A   2  ( 0 . 85 Hz)  (18 . 0  10  2 m)  5.13 m/s 2 . v max 
2
ωA  2 πfA  0 . 961 m/s
v  ( 2 πf )
. b) a x   ( 2 πf ) 2 x   2 . 57 m/s 2 ,
A  x
2
  2 π ( 0 . 85 Hz)
2

(18 . 0  10
2
m)
2
 ( 9 . 0  10
2
m)
2
 0 . 833 m/s.
c) The fraction of one period is (1 2 π ) arcsin (12 . 0 18 . 0 ), and so the time is
1
(T 2 π )  arcsin (12 . 0 18 . 0 )  1 . 37  10 s. Note that this is also arcsin ( x A ) ω .
d) The conservation of energy equation can be written 12 kA 2  12 mv 2  12 kx 2 . We are
given amplitude, frequency in Hz, and various values of x . We could calculate velocity
from this information if we use the relationship k m  ω 2  4 π 2 f 2 and rewrite the
conservation equation as
1
2
A 
2
2
v
1
2 4π 2 f
2

1
2
x
2
. Using energy principles is generally a good
approach when we are dealing with velocities and positions as opposed to accelerations
and time when using dynamics is often easier.
13.26: In the example, A2  A1
m  3M
heat.
M
M m
and now we want A 2 
. For the energy, E 2  12 kA 22 , but since A 2 
1
2
1
2
A1 , E 2 
A1 . So
1
4
1
2
E 1 , or

3
4
M
M m
, or
E 1 is lost to
13.27: a)
1
2
mv
2

1
2
kx
2
 0 . 0284 J
2
b)
x0 
2
v0
ω
c) ωA 

2
( 0 . 012 m)
2
.
2
( 0 . 300 m/s)

 0 . 014 m.
( 300 N/m) (0.150 kg)
k mA  0 . 615 m s 
13.28: At the time in question we have
x  A cos ( ωt   )  0 . 600 m
v   ωA sin( ωt   )  2 . 20 m s
a   ω A cos ( ωt   )   8 . 40 m s
2
2
Using the displacement and acceleration equations:
 ω A cos ( ωt   )   ω ( 0 . 600 m)   8 . 40 m s
2
2
ω  14 . 0 and ω  3 . 742 s
2
1
2
To find A, multiply the velocity equation by ω :
1
 ω A sin ( ωt   )  ( 3 . 742 s ) (2.20 m s )  8 . 232 m s
2
2
Next square both this new equation and the acceleration equation and add them:
ω A sin ( ωt   )  ω A cos ( ωt   )  ( 8 . 232 m s )  (  8 . 40 m s )
4
2
2
4
2
2
2
2
2
2
 ω A sin ( ωt   )  cos ( ωt   )
4
2
2
2
ω A  67 . 77 m
2
s  70 . 56 m
138 . 3 m
2
s
4
2
A 
2
ω
4
4
4

138 . 3 m
2
s  138 . 3 m
2
s
4
2
s
4
4
1 4
 0 . 7054 m
2
( 3 . 742 s )
A  0 . 840 m
The object will therefore travel 0 . 840 m  0 . 600 m  0.240 m to the right before stopping
at its maximum amplitude.
13.29: v max  A k m
Use T to find k m :
T  2π
m k so k m  ( 2 π T )  158 s
2
2
Use a max to find A :
a max  kA m so A  a max ( k m )  0 . 0405 m.
Then v max  A k m  0 . 509 m s
13.30:
Using k 
F0
m 
13.31: a) k 
mg
Δl
from the calibration data,
L0
( F0 L 0 )
( 2 πf )
2
( 200 N) (1.25  10

1
m)
2
(2 π ( 2 . 60 Hz))
 6 . 00 kg.
2

b) T  2 π 
( 650 kg) (9.80 m s )
 53 1  10 N m.
3
( 0 . 120 m)
m
l
 2π
k
0 . 120 m
 2π
g
9.80 m s
2
 0 . 695 s.
13.32: a) At the top of the motion, the spring is unstretched and so has no potential
energy, the cat is not moving and so has no kinetic energy, and the gravitational potential
energy relative to the bottom is 2 mgA  2 ( 4 .00 kg)(9.80 m/s 2 )  ( 0 .050 m)  3.92 J .
This is the total energy, and is the same total for each part.
b) U grav  0 , K  0 , so U spring  3 . 92 J .
c) At equilibrium the spring is stretched half as much as it was for part (a), and so
U spring  14 ( 3 . 92 J)  0.98 J, U grav  12 ( 3 . 92 J)  1.96 J, and so K  0 . 98 J .
13.33: The elongation is the weight divided by the spring constant,
l 
w
k

mg
2
ω m

gT
2
4π
2
 3 . 97 cm .
13.34: See Exercise 9.40. a) The mass would decrease by a factor of (1 3 ) 3  1 27 and so
2
the moment of inertia would decrease by a factor of (1 27 )( 1 3 )  (1 243 ) , and for the
same spring constant, the frequency and angular frequency would increase by a factor of
243  15 . 6 . b) The torsion constant would need to be decreased by a factor of 243, or
changed by a factor of 0.00412 (approximately).
13.35: a) With the approximations given, I  mR 2  2 . 72  10  8 kg  m 2 ,
8
2
or 2.7  10 kg  m to two figures.
b) κ  ( 2 πf ) 2 I  ( 2 π 2 Hz) 2 ( 2 . 72  10  8 kg  m 2 )  4 .3  10  6 N  m rad .
13.36: Solving Eq. (13.24) for  in terms of the period,
2
 2π 
 
 I
 T 
2
 2π 
3
2
2

 (( 1 2 )( 2 . 00  10 kg)(2.20  10 m) )
 1 . 00 s 
 1 . 91  10
5
N  m/rad.
13.37:

I 
( 2 πf )
2

0 . 450 N  m/rad
2 π (125)
 0 . 0152 kg  m .
2
(265 s) 
2
13.38: The equation θ   cos ( ω t  φ ) describes angular SHM. In this problem, φ  0 .
dθ
dt
a)
  ω  sin( ω t ) and
2
d θ
dt
2
  ω  cos( ω t ).
2
b) When the angular displacement is  ,    cos( ω t ) , and this occurs at t  0 , so
dθ
2
 0 since
sin(0)  0, and
d θ
dt
dt
  ω  , since cos(0)  1.
2
2
When the angular displacement is  2 , 2   cos( ω t ), or
dθ

 ω 3
dt
since sin( ω t ) 
3
2
, and
2
2
This corresponds to a displacement of 60  .
 cos( ω t ).
2
d θ
dt
ω 
1
2

2
, since cos( ω t )  1 2 .
2
13.39: Using the same procedure used to obtain Eq. (13.29), the potential may be
expressed as
U  U 0 [( 1  x R 0 )
 12
6
 2 (1  x R 0 ) ].
Note that at r  R 0 , U  U 0 . Using the appropriate forms of the binomial theorem for
| x R 0 | << 1,


  12   13 
 x R 0 2  
  1  12  x R 0  
2


U  U0


  6   7 

 x R 0 2  
  21  6 x R0  
2




36 2 
 U 0  1  2 x 
R0



1
2
kx  U 0 .
2
where k  72 U 0 / R 2 has been used. Note that terms in u 2 from Eq. (13.28) must be
kept ; the fact that the first-order terms vanish is another indication that R 0 is an extreme
(in this case a minimum) of U.
13.40: f 
1
k
2
m 2 

1
2
2 ( 580 N/m)
1 . 008 (1 . 66  10  27
 1 . 33  10
14
Hz.
kg)
13.41: T  2 L g , so for a different acceleration due to gravity g ,
T  T
g g   1 . 60 s  9 . 80 m s
2
3 . 71 m s  2 . 60 s.
2
13.42: a) To the given precision, the small-angle approximation is valid. The highest
speed is at the bottom of the arc, which occurs after a quarter period, T4  2 Lg  0 . 25 s.
b) The same as calculated in (a), 0.25 s. The period is independent of amplitude.
13.43: Besides approximating the pendulum motion as SHM, assume that the angle
is sufficiently small that the length of the spring does not change while swinging in the
arc. Denote the angular frequency of the vertical motion as 0 

1
2
ω0 
kg
4w
,
k
m

kg

and  
g
L
which is solved for L  4 w k . But L is the length of the stretched
spring; the unstretched length is L0  L  w k  3 w k  31 . 00 N  1 .50 N/m   2 . 00 m.
13.44:
13.45: The period of the pendulum is T  136 s  100  1 . 36 s. Then,
4π L
T
2
4 π . 5 m 
2
2
g 

1.36 s 
2
 10 . 67 m s .
2
13.46: From the parallel axis theorem, the moment of inertia of the hoop about the nail is
2
2
2
I  MR  MR  2 MR , so T  2 π 2 R g , with d  R in Eq. 13.39 . Solving for R,
R  gT
2
2
8 π  0 . 496 m.
13.47: For the situation described, I  mL 2 and d  L in Eq. (13.39); canceling the factor
of m and one factor of L in the square root gives Eq. (13.34).
13.48:
a) Solving Eq. (13.39) for I,
2
 T 
 0 . 940 s 
I 
 mgd  

 2π 
 2π 
2
1.80

kg  9 . 80 m s
2
 0 .250 m   0 .0987
kg  m .
2
b) The small-angle approximation will not give three-figure accuracy for
From energy considerations,
Θ  0.400 rad.
mgd 1  cos   
1
2
2
I Ω max .
Expressing  max in terms of the period of small-angle oscillations, this becomes
2
 2π 
2
 1  cos   
 T 
 max 
2
 2π 
2
 1  cos 0.40 rad
 0 . 940 s 
  2 . 66
rad s .
13.49: Using the given expression for I in Eq. (13.39), with d=R (and of course m=M),
T  2 π 5 R 3 g  0 . 58 s.
13.50: From Eq. (13.39),
2
 T 
2
I  mgd 
  1 . 80 kg  9.80 m s
π
2




2
120 s 100 
2
0 . 200 m  
  0 . 129 kg.m .
π
2


13.51: a) From Eq. (13.43),
2 . 50 N m  0 . 90 kg s 2

 2 . 47 rad
0 . 300 kg  4 0 . 300 kg 2
km  2  2 . 50 N m  0 . 300 kg   1 . 73 kg
ω 
b) b  2
13.52: From Eq. (13.42) A2  A1 exp 
b 
2m
t
b
2m
t . Solving
s , so f  
ω
2π
 0 . 393 Hz.
s.
for b ,
 A  2 ( 0 . 050 kg )  0 . 300 m 
  0 . 0220 kg s.
ln  1  
ln 
( 5 . 00 s )
 0 . 100 m 
 A2 
As a check, note that the oscillation frequency is the same as the undamped frequency to
3
4 . 8  10 % , so Eq. (13.42) is valid.
13.53: a) With   0, x (0)  A.
b)
vx 
dx
 Ae
 (b 2 m )t
dt
b





  2 m cos ω t  ω sin ω t  ,


and at t  0 , v   Ab 2 m ; the graph of x versus t near t  0 slopes down.
c)
ax 
dv x
 Ae
 (b 2 m )t
dt
 b 2

ω b
2 
 ω   cos  ' t 
sin ω t  ,
 
2
2m

 4 m

and at t  0 ,
 b2
 b2
k 
2 



.
a x  A 

ω

A

2
2

m 
 4m

 2m
(Note that this is (  bv 0  kx 0 ) m .) This will be negative if
b
2 km , zero if b 
2 km and positive
if b 
2 km .
The graph in the three cases will
be curved down, not curved, or curved up, respectively.
13.54: At resonance, Eq. (13.46) reduces to A  Fmax b  d . a) 3 . b) 2 A1 . Note that
the resonance frequency is independent of the value of b (see Fig. (13.27)).
A1
13.55: a) The damping constant has the same units as force divided by speed, or
kg  m s 2  m s   kg s .  b)The units of km are the same as [[ kg s 2 ][ kg]] 1 2  kg s ,
the same as those for b . c) ω d2  k m . (i) bω d  0 . 2 k , so A  Fmax 0 . 2 k   5 Fmax k .
( ii) b d  0.4 k , so A  Fmax ( 0 . 4 k )  2 . 5 Fmax k , as shown in Fig.(13.27).
13.56: The resonant frequency is
k m 
( 2.1  10 N m ) 108 kg )  139 rad s  22.2 Hz,
6
and this package does not meet the criterion.
13.57: a)
a  Aω
2
 π
rad s  
 0 . 100 m  
 

  ( 3500 rev min ) 


2
 30 rev min  
2
2
 6 . 72  10 m s .
3
 π
rad s 
  18 . 3 m s .
 30 rev min 
b) ma  3 . 02  10 3 N. c) ωA  ( 3500 rev min ) (. 05 m ) 
K 
1
2
mv
2
 ( 12 )(. 45 kg )( 18 . 3 m s )  75 . 6 J.
2
d) At the midpoint of the stroke, cos( ω t)=0
and so ωt  π 2 , thus t  π 2 ω. ω  (3500 rev min )( 30π
t 
3
2 ( 350 )
rad s
rev min
)
350 π
3
rad s,
so
3
s. Then P   K  t , or P  75 . 6 J ( 2(350)
s)  1.76  10 W.
4
e) If the frequency doubles, the acceleration and hence the needed force will quadruple
(12.1  10 3 N). The maximum speed increases by a factor of 2 since v α ω , so the speed
will be 36.7 m/s. Because the kinetic energy depends on the square of the velocity, the
kinetic energy will increase by a factor of four (302 J). But, because the time to reach the
midpoint is halved, due to the doubled velocity, the power increases by a factor of eight
(141 kW).
13.58: Denote the mass of the passengers by m and the (unknown) mass of the car by M.
The spring cosntant is then k  mg  l . The period of oscillation of the empty car is
TE  2 π
M k
and the period of the loaded car is
M m
TL  2 π
TE 

k
TL  2 π 
2
2
l
TE  2 π 
2
2
l
, so
g
 1 . 003 s.
g
13.59: a) For SHM, the period, frequency and angular frequency are independent of
amplitude, and are not changed. b) From Eq. (13.31), the energy is decreased by a factor
of 14 . c) From Eq. (13.23), the maximum speed is decreased by a factor of 12 d) Initially,
the speed at A1 4 was
ω
 A1
2    A1 4  
2
2
15
4
3
4
ωA 1 ;
ωA 1 ,
after the amplitude is reduced, the speed is
so the speed is decreased by a factor of
1
5
(this result is
valid at x   A1 4 as well). e) The potential energy depends on position and is
unchanged. From the result of part (d), the kinetic energy is decreased by a factor of
13.60: This distance L is L  mg k ; the period of the oscillatory motion is
T  2π
m
k
 2
L
,
g
which is the period of oscillation of a simple pendulum of lentgh L.
1
5
.
13.61: a) Rewriting Eq. (13.22) in terms of the period and solving,
T 
2π
A  x
2
2
 1 . 68 s.
v
b) Using the result of part (a),
x 
 vT 
2
A 

 2 
2
 0 . 0904 m.
c) If the block is just on the verge of slipping, the friction force is its maximum,
2
2
f  μ s n  μ s mg . Setting this equal to ma  mA  2 π T  gives μ s  A  2 π T  g  0 . 143 .
13.62: a) The normal force on the cowboy must always be upward if he is not holding on.
He leaves the saddle when the normal force goes to zero (that is, when he is no longer in
contact with the saddle, and the contact force vanishes). At this point the cowboy is in
free fall, and so his acceleration is  g ; this must have been the acceleration just before
he left contact with the saddle, and so this is also the saddle’s acceleration.
b) x   a ( 2 π f ) 2   ( 9 . 80 m s 2 ) 2 π (1 . 50 Hz )) 2  0 .110 m. c) The cowboy’s speed will
be the saddle’s speed, v  ( 2 πf ) A 2  x 2  2 . 11 m s. d) Taking t  0 at the time when
the cowboy leaves, the position of the saddle as a function of time is given by Eq.
(13.13), with cos   
x
g
ω
2

a
ω
2
g
2
ω A
;
this is checked by setting t  0 and finding that
. The cowboy’s position is x c  x 0  v 0 t  ( g 2 ) t . Finding the time at which
2
the cowboy and the saddle are again in contact involves a transcendental equation which
must be solved numerically; specifically,
( 0 . 110 m)  ( 2 . 11 m s) t  ( 4 . 90 m s ) t  ( 0 . 25 m) cos (( 9 . 42 rad s) t  1 . 11 rad),
2
2
which has as its least non-zero solution t  0 . 538 s. e) The speed of the saddle is
2
(  2 . 36 m s) sin ( ω t   )  1 . 72 m s , and the cowboy’s speed is (2.11 m s )  ( 9 . 80 m s )
 ( 0 . 538 s)   3 . 16 m s, giving a relative speed of 4 . 87 m s (extra figures were kept in
the intermediate calculations).
13.63: The maximum acceleration of both blocks, assuming that the top block does not
slip, is a max  kA ( m  M ), and so the maximum force on the top block is
 m mM  kA 
μ s mg , and so the maximum
amplitude
is Amax  μ s ( m  M ) g k .
13.64: (a) Momentum conservation during the collision: mv 0  ( 2 m )V
1
V 
2
v0 
1
( 2 . 00 m s)  1 . 00 m s
2
Energy conservation after the collision:
1
MV
2
2
MV
x

T 
1
1
f

kx
2
2
 0 . 500 m (amplitude
k M
k M 
2π
1
80 . 0 N m
ω  2 πf 
f 

2
( 20 . 0 kg)( 1 . 00 m s)
k
2
1
80 . 0 N m
2π
20 . 0 kg
1
 0 . 318 Hz
 3 . 14 s
0 . 318 Hz
(b) It takes 1 2 period to first return: 12 ( 3 . 14 s)  1 . 57 s
)
a) m  m 2
13.65:
Splits at x  0 where energy is all kinetic energy, E  12 mv 2 , so E  E 2
k stays same
E 
kA so A 
2
1
2
2E k
Then E  E 2 means A  A
T  2π
2
m k so m  m 2 means T  T
2
b) m  m 2
Splits at x  A where all the energy is potential energy in the spring, so E doesn’t
change.
2
E  12 kA so A stays the same.
T  2π
m k so T  T
as in part (a).
c) In example 13.5, the mass increased. This means that T increases rather than
decreases. When the mass is added at x  0 , the energy and amplitude change. When the
mass is added at x   A , the energy and amplitude remain the same. This is the same as
in this problem.
2,
13.66: a)
For space considerations, this figure is not precisely to the scale suggested in the
problem. The following answers are found algebraically, to be used as a check on the
graphical method.
c)
v0  
2E
A
b)
2 ( 0 . 200 J)

k
E
4
 0 . 050 J. d) If U 
2K0
m
and x 0 
2U 0
k
1
2
E, x 

0.429
2
 0 . 141 m.
e) From Eq. (13.18), using
,

and   arctan
A
 0 . 200 m.
(10.0 N/m)
v0
ω x0
  0 .580 rad .
2K0

m
k
m
2U 0
k

K0
U0

0 . 429
13.67: a) The quantity  l is the amount that the origin of coordinates has been moved
from the unstretched length of the spring, so the spring is stretched a distance  l  x (see
Fig. (13.16 ( c ))) and the elastic potential energy is U el  (1 2 ) k (  l  x ) 2 .
U  U el  mg  x  x 0  
b)
1
kx 
2
2
1
2
 l  2
 k  lx  mgx  mgx 0 .
Since  l  mg k , the two terms proportional to x cancel, and
U 
1
kx 
2
2
1
2
k   l   mgx 0 .
2
c) An additive constant to the mechanical energy does not change the dependence
, and so the relations expressing Newton’s laws and the
of the force on x , F x   dU
dx
resulting equations of motion are unchanged.
13.68:
The “spring constant” for this wire is k 
f 
1
k
2π
m

1
g
2π
l

1
2π
mg
l
,
so
9 . 80 m s
2 . 00  10
3
2
 11 . 1 Hz.
m
a) 2TπA  0 . 150 m s. b) a    2 π T  x   0 . 112 m s 2 . The time to go from
equilibrium to half the amplitude is sin ωt  1 2 , or ωt  π 6 rad, or one-twelfth of a
period. The needed time is twice this, or one-sixth of a period, 0.70 s.
d)  l  mgk  ωg   2 πg r   4 . 38 m.
2
13.69:
2
2
13.70: Expressing Eq. (13.13) in terms of the frequency, and with   0 , and taking
two derivatives,
 2 πt 

x  0 . 240 m  cos 
 1 . 50 s 
 2 π 0 . 240 m  
 sin
v x   
1 . 50 s  

 2 πt 

   1 . 00530 m s  sin
 1 . 50 s 
2
 2 πt 


 1 . 50 s 
 2π 
 2 πt 
 2 πt 
2
  0 . 240 m  cos 
   4 . 2110 m s cos 
.
a x   
 1 . 50 s 
 1 . 50 s 
 1 . 50 s 


a) Substitution gives x   0 . 120 m, or using t 
b) Substitution gives
ma x   0 . 0200 kg 2 . 106 m s
c) t 
T
2π
arccos

3 A 4
A
  0 . 577
2
  4 .21  10
2
T
3
gives x  A cos 120  
A
2
.
N, in the  x - direction.
s.
d) Using the time found in part (c) , v  0 . 665 m s (Eq.(13.22) of course gives the
same result).
13.71: a) For the totally inelastic collision, the final speed v in terms of the initial
speed V  2 gh is
v V
M
mM


2 9 . 80 m s
the steak hits, the pan is
2
v0
ω
2
is v
2
 k  m  M  
2
0 .40 m     2 .57
Mg
k
2 ghM
2 .2
2 .4

to two figures. b) When
above the new equilibrium position. The ratio
k  m  M ,
2
2
A
m s, or 2.6 m s
and so the amplitude of oscillation is
2
2 ghM
 Mg 

 
k m  M
 k 

  2 . 2 kg  9 . 80 m/s

 400 N m 


2
 
2
 

2
2 ( 9 . 80 m s )(0.40 m)(2.2 kg)
2
( 400 N m)(2.4 kg)
 0 . 206 m.
(This avoids the intermediate calculation of the speed.) c) Using the total mass,
T  2π
( m  M ) k  0 . 487 s.
13.72: f  0 . 600 Hz, m  400 kg; f  12 mk gives k  5685 N/m.
This is the effective force constant of the two springs.
a) After the gravel sack falls off, the remaining mass attached to the springs is 225
kg. The force constant of the springs is unaffected, so f  0 . 800 Hz.
To find the new amplitude use energy considerations to find the distance
downward that the beam travels after the gravel falls off.
Before the sack falls off, the amount x 0 that the spring is stretched at equilibrium is
given by mg  kx 0 , so x 0  mg k   400 kg  9 . 80 m/s 2  5685 N/m   0 . 6895 m. The
maximum upward displacement of the beam is A  0 . 400 m. above this point, so at this
point the spring is stretched 0.2895 m.
With the new mass, the mass 225 kg of the beam alone, at equilibrium the spring is
stretched mg k  ( 22 5 kg) (9.80 m s 2 ) ( 5685 N m)  0.6895 m. The new amplitude is
therefore 0 . 3879 m  0 . 2895 m  0 . 098 m. The beam moves 0.098 m above and below the
new equilibrium position. Energy calculations show that v  0 when the beam is 0.098 m
above and below the equilibrium point.
b) The remaining mass and the spring constant is the same in part (a), so the new
frequency is again 0 . 800 Hz.
The sack falls off when the spring is stretched 0.6895 m. And the speed of the beam
at this point is v  A k m  0 . 400 m   5685 N/m   400 kg   1 . 508 m/s. . Take y  0
at this point. The total energy of the beam at this point, just after the sack falls off, is
2
2
E  K  U el  U g  12  225 kg 1 . 508 m/s   12 5695 N/m 0 . 6895 m   0  1608 J. Let this
be point 1. Let point 2 be where the beam has moved upward a distance d and where
2
v  0 . E 2  12 k 0 . 6985 m  d   mgd . E 1  E 2 gives d  0 . 7275 m . At this end point of
motion the spring is compressed 0.7275 m – 0.6895 m =0.0380 m. At the new
equilibrium position the spring is stretched 0.3879 m, so the new amplitude is 0.3789 m +
0.0380 m = 0.426 m. Energy calculations show that v is also zero when the beam is 0.426
m below the equilibrium position.
13.73: The pendulum swings through
Use T to find g:
T  2π
L g so g  L  2 π T
2
1
2
cycle in 1.42 s, so T  2 . 84 s. L  1 . 85 m.
 9 . 055 m/s
Use g to find the mass M p of Newtonia:
g  GM
2
p
/ Rp
2 πR p  5 . 14  10 m, so R p  8 . 18  10 m
7
6
2
mp 
gR p
G
 9 . 08  10
24
kg
2
13.74:
a) Solving Eq. (13.12) for m , and using k 
2
F
l
2
 T  F
 1  40 . 0 N
m 

 4 . 05 kg.


 2   l  2  0.250 m
b) t  ( 0 .35 )T , and so x   A sin2 π (0.35)   0.0405 m. Since t  T4 , the mass has
already passed the lowest point of its motion, and is on the way up.
c) Taking upward forces to be positive, F spring  mg   kx , where x is the
displacement from equilibrium , so
Fspring   (160 N m)(  0.030 m)  (4.05 kg)(9.80
m/s )  44 . 5 N.
2
13.75:
Of the many ways to find the time interval, a convenient method is to take
  0 in Eq. (13.13) and find that for x  A 2 , cos ω t  cos( 2 πt / T )  21 and so t  T / 6 .
The time interval available is from  t to t , and T / 3  1 . 17 s.
13.76:
See Problem 12.84; using x as the variable instead of r ,
F ( x)  
dU

dx
GM E m
R
3
E
x , so ω 
2
GM
R
3
E
E

g
RE
The period is then
T 
or 84.5 min.
2π
ω
 2π
RE
g
6 . 38  10 m
6
 2π
9.80 m/s
2
 5070 s,
.
13.77:
Take only the positive root (to get the least time), so that
dx

dt
dx
A  x
2

A
0
2
dx
A  x
2
2
k

k
2
dt
m

k
m
t1
0
k
m
2
2
m
arcsin( 1) 
π
A  x , or

k
m
dt 
k
m
( t1 )
t1
t1 ,
where the integral was taken from Appendix C. The above may be rearranged to show
that t1   k  T4 , which is expected.
2
m
x
13.78:
a)
x
U    F dx  c  x dx 
3
0
0
c
4
x .
4
a) From conservation of energy, 12 mv 2  4c ( A 4  x 4 ) , and using the technique of Problem
13.77, the separated equation is
dx
A  x
4
c

dt .
2m
4
Integrating from 0 to A with respect to x and from 0 to T 4 with respect to t ,
A

0
dx
A  x
4
4
c T

.
2m 4
To use the hint, let u  Ax , so that dx  a du and the upper limit of the u  integral is
u  1.
Factoring A 2 out of the square root,
1
1
A
0
which may be expressed as T 
motion is not simple harmonic.
7 . 41
A
du
1 u
m
c
.
4

1 . 31

A
c
T,
32 m
c) The period does depend on amplitude, and the
13.79: As shown in Fig. 13 .5 b , v   v tan sin θ . With v tan  Aω and θ  ωt   , this is
Eq. 13 . 15 .
13.80:
a) Taking positive displacements and forces to be upwad,
2
n  mg  ma , a    2 πf  x , so

n  m g   2 πf
2 A cos 2 πf  t   .
a) The fact that the ball bounces means that the ball is no longer in contact with the lens,
and that the normal force goes to zero periodically. This occurs when the amplitude
of the acceleration is equal to g , or when
g   2 πf b  A .
2
13.81: a) For the center of mass to be at rest, the total momentum must be zero, so the
momentum vectors must be of equal magnitude but opposite directions, and the momenta


can be represented as p and  p .
K tot  2 
b)
p
2
2m

p
2
2 m 2 
.
c) The argument of part (a) is valid for any masses. The kinetic energy is
K tot 
p
2
2m1

p
2
2m 2

2
p  m1  m 2

2  m 1 m 2
2

p

 2  m m  m  m  .
1
2
1
2

Fr  
13.82: a)
 R 7 
1 
 Α   90   2  .
dr
 r  r 
dU
b) Setting the above expression for F r equal to zero, the term in square brackets
7
vanishes, so that
R0
r
9

1
r
2
, or R 0  r , and r  R 0 .
7
7
U R0   
c)
7Α
  7 . 57  10
 19
J.
8 R0
d) The above expression for F r can be expressed as
9
2
 r  
A  r 
 

Fr  2  
R  
R 0   R 0 

0 



A
R

2
0
A
2
R0

A
2
R0
1   x
R 0 
1  9  x
 7 x
9
 1   x R 0 
2

R 0   1  2  x R 0 
R0 
 7A
   3  x .
 R0 
e)
f 
1
2π
k m 
1
7A
2π
R0 m
3
 8 . 39  10  z.
12
13.83: a)
Fr  
 1
1
 A 2 
dx
r  2 R 0  2
 r
dU

.

b) Setting the term in square brackets equal to zero, and ignoring solutions with
r  0 or r  2 R 0 , r  2 R 0  r , or r  R 0 .
c) The above expression for F r may be written as
2
2
 r
 
A  r 
 

F r  2  
 R  2 
R 0   R 0 
0

 


A
R

2
0
A
2
R0
1   x
R 0 
1  2  x
2
 1   x R 0 
2

R 0   1    2   x R 0 
 4A 
   3  x ,
 R0 
3
corresponding to a force constant of k  4 A R 0 . d) The frequency of small oscillations
3
would be f  (1 2 π ) k m  (1 π ) A mR 0 .
13.84: a) As the mass approaches the origin, the motion is that of a mass attached to a
spring of spring constant k, and the time to reach the origin is π2 m k . After passing
through the origin, the motion is that of a mass attached to a spring of spring constant 2k
and the time it takes to reach the other extreme of the motions is π2 m 2 k . The period is
twice the sum of these times, or T  π
m
k
1   . The period does not depend on the
1
2
amplitude, but the motion is not simple harmonic. B) From conservation of energy, if the
negative extreme is A , 12 kA 2  12 ( 2 k ) A  2 , so A    A2 ; the motion is not symmetric about
the origin.
13.85: There are many equivalent ways to find the period of this oscillation. Energy
considerations give an elegant result. Using the force and torque equations, taking torques
about the contact point, saves a few intermediate steps. Following the hint, take torques
about the cylinder axis, with positive torques counterclockwise; the direction of positive
rotation is then such that   Ra , and the friction force f that causes this torque acts in
the –x-direction. The equations to solve are then
fR  I cm ,
Ma x   f  kx ,
a  R ,
Which are solved for
ax 
kx
M  I R
2
 
k
x,
(3 2 ) M
where I  I cm  (1 2 ) MR 2 has been used for the combination of cylinders. Comparison
with Eq. (13.8) gives T 
2π
ω
 2π
3M 2k .
13.86: Energy conservation during downward swing:
m 2 gh 0 
1
2
v
m 2v
2
2 gh 0 
2
2 ( 9 . 8 m s )( 0 . 100 m )  1 . 40 m s
Momentum conservation during collision:
m 2 v  ( m 2  m 3 )V
V 
m 2v
m2  m3

( 2 . 00 kg )( 1 . 40 m s )
 0 . 560 m s
5 . 00 kg
Energy conservation during upward swing:
Mgh f 
1
hf  V
2g 
2
MV
2
2
( 0 . 560 m s )
2
2
 0 . 0160 m  1.60 cm
2 ( 9 . 80 m s )
cos θ 
48 . 4 cm
50.0 cm
θ  14 . 5 
f 
1
g
2
l

1
9 . 80 m s
2
0 . 500 m
2
 0 . 705 Hz
13.87:
T  2π
I mgd , m  3 M
d  y cg 
d 
m 1 y1  m 2 y 2
m1  m 2
2 M 1 . 55 m  2   M 1 . 55 m  1.55 m  2 
 1 . 292 m
3M
I  I1  I 2
I1 
1
3
 2 M 1 . 55
I 2 , cm 
1
12
m   1 . 602 m
M 1 . 55 m 
2
2
M
2
The parallel-axis theorem (Eq. 9.19) gives

I 2  I 2 , cm  M 1 . 55 m  1 . 55 m  2   5 . 06 m
2
I  I 1  I 2  7 . 208 m
2
2
M
M
7 .208 m  M
 2 . 74 s.
3 M 9 . 80 m s 1 . 292 m 
2
Then T  2 π I mgd  2 π
2
This is smaller than T  2 . 9 s found in Example 13.10.
13.88: The torque on the rod about the pivot (with angles positive in the direction
indicated in the figure) is τ   k L2 θ  L2 . Setting this equal to the rate of change of angular
momentum, I   I
d 
2
dt
2
,
2
d θ
dt
2
2
 k
L 4
θ 
I
3k
θ,
M
where the moment of inertia for a slender rod about its center, I 
It follows that ω 2 
3K
M
, and T 
2π
ω
 2π
M
3k
1
12
ML
2
has been used.
.
13.89: The period of the simple pendulum (the clapper) must be the same as that of the
bell; equating the expression in Eq. (13.34) to that in Eq. (13.39) and solving for L gives
2
L  Ι md  (18 . 0 kg  m ) (( 34 . 0 kg)( 0 . 60 m))  0 . 882 m. Note that the mass of the bell,
not the clapper, is used. As with any simple pendulum, the period of small oscillations of
the clapper is independent of its mass.
13.90: The moment of inertia about the pivot is 2 (1 3 ) ML 2  ( 2 3 ) ML 2 , and the center of
gravity when balanced is a distance d  L (2 2 ) below the pivot (see Problem 8.95).
From Eq. (13.39), the frequency is
f 
1
T

1
3g
2π
4 2L

1
4π
3g

2L
13.91: a) L  g (T 2 π ) 2  3 .97 m. b)There are many possibilities. One is to have a
uniform thin rod pivoted about an axis perpendicular to the rod a distance d from its
center. Using the desired period in Eq. (13.39) gives a quadratic in d, and using the
maximum size for the length of the rod gives a pivot point a distance of 5.25 mm, which
is on the edge of practicality. Using a “dumbbell,” two spheres separated by a light rod of
length L gives a slight improvement to d=1.6 cm (neglecting the radii of the spheres in
comparison to the length of the rod; see Problem 13.94).
13.92: Using the notation 2bm  γ , mk  ω 2 and taking derivatives of Eq. (13.42) (setting the
phase angle   0 does not affect the result),
x  Αe
 t
cos ω  t
vx   Αe
 t
( ω  sin ω  t   cos ω  t )
ax   Αe
 t
(( ω   γ ) cos ω  t  2 ω   sin ω  t ).
2
2
Using these expression in the left side of Eq. (13.41),
 kx  bv x  Αe
 t
(  k cos ω  t  ( 2  m ) ω  sin ω  t  2 m γ cos ω  t )
 mΑ e
2
 t
(( 2 γ  ω ) cos ω  t  2 γ ω  sin ω  t ).
2
2
The factor ( 2 γ 2  ω 2 ) is γ 2  ω  2 (this is Eq. (13.43)), and so
 kx  bv x  m Α e
 t
(( γ  ω  ) cos ω  t  2 γ ω  sin ω  t )  ma x .
2
2
13.93: a) In Eq. (13.38), d=x and from the parallel axis theorem,
gx
2
2
2
2
I  m ( L 12  x ) , so ω 
. b) Differentiating the ratio ω g 
( L 12 )  x
2
2
x
(L
2
12 )  x
2
with
respect to x and setting the result equal to zero gives
1
( L 12 )  x
2
Which is solved for x  L
2

2x
2
, or 2 x  x  L 12 ,
2
(( L 12 )  x )
2
2
2
3g
ω
2
2
12 .
c) When x is the value that maximizes ω the ratio
so the length is L 
2
 0 . 430 m.
ω
g
2

L

12
2
2 L 12


6
L 12

3
L,
13.94: a) From the parellel axis theorem, the moment of inertia about the pivot point
2
2
is
M L   2 5 R .
Using this in Eq. (13.39), With d  L gives.
L  2 5 R
2
T  2π
2
 2π
gL
b) Letting
L
1  2R
2
g
1 2R
2
5 L  1 . 001
2
square root as 1  R 5 L ) gives
c) 14 . 1  1 . 270 cm   18 . 0 cm .
2
2
L
R
5 L  Tsp 1  2 R
2
2
2
5L .
and solving for the ratio L R (or approximating the
 14 . 1 .
13.95: a) The net force on the block at equilibrium is zero, and so one spring (the one
with k 1  2 . 00  m ) must be stretched three times as much as the one with
k 2  6 . 00  m . The sum of the elongations is 0.200 m, and so one spring stretches 0.150
m and the other stretches 0.050 m, and so the equilibrium lengths are 0.350 m and 0.250
m. b) There are many ways to approach this problem, all of which of course lead to the
result of Problem 13.96(b). The most direct way is to let  x1  0 . 150 m and
x 2  0 . 050 m, the results of part (a). When the block in Fig.(13.35) is displaced a
distance x to the right, the net force on the block is
 k 1   x 1  x   k 2   x 2  x   k 1  x 1  k 2  x 2    k 1  k 2  x .
From the result of part (a), the term in square brackets is zero, and so the net force is
  k 1  k 2  x , the effective spring constant is k eff  k 1  k 2 and the period of vibration is
T  2π
0 . 100 kg
8.00  m
 0 . 702 s.
13.96: In each situation, imagine the mass moves a distance  x , the springs move
distances  x 1 and  x 2 , with forces F1   k 1  x1 , F 2   k 2  x 2 .
a)  x   x1   x 2 , F  F1  F2   k1  k 2  x , so k eff  k1  k 2 .
b) Despite the orientation of the springs, and the fact that one will be compressed when
the other is extended,  x   x1   x 2 , and the above result is still valid; k eff  k 1  k 2 .
c) For massless springs, the force on the block must be equal to the tension in any point
of the spring combination, and F  F1  F 2 , and so  x1  
F
k1
,  x2  
F
k2
, and
 1
k  k2
1 
F   1
 x   

F

k1k 2
 k1 k 2 
and κ eff 
κ1κ 2
κ1  κ 2
. d) The result of part (c) shows that when a spring is cut in half, the
effective spring constant doubles, and so the frequency increases by a factor of
2.
13.97: a) Using the hint,
T   T  2π
g
 1 2 1  3 2

L g
 g
g  T  T
,


2
2g
so  T   1 2 T g  g . This result can also be obtained from T 2 g  4 π 2 L , from which
 2T  T  g  T 2  g

 0 . Therefore,
and g   g  g  1 

T
T

2T 
2
  9 . 80 m s
T 


1 g
2 g
.
b) The clock runs slow;  T  0 ,  g  0

2  4 . 00 s  
2
 1 
  9 . 7991 m s .


86
,
400
s


13.98: Denote the position of a piece of the spring by l ; l  0 is the fixed point and
l  L is the moving end of the spring. Then the velocity of the point corresponding to l ,
l
denoted u , is u l   v
L
(when the spring is moving, l will be a function of time, and so
u is an implicit function of time). a) dm 
dK 
1
M
L
dl , and so
dm u 
2
2
1 Mv
2
2
l dl ,
3
2 L
and
K 
b) mv
dv
dt
ω 
3k
M
 kx
dx
dt

dK 
2 L
Mv
3
2L
l
2
dl 
Mv
2
.
6
0
 0 , or ma  kx  0 , which is Eq. 13 . 4 . c) m is replaced by
and M  
M
3
M
3
, so
.
13.99: a) With I  1 3 ML 2 and d  L 2 in Eq. 13 . 39 , T 0  2 π 2 L 3 g . With the
addedmass, I  M L2 3   y 2 , m  2 M and d   L 4   y 2 , T  2 π

L
2
3 y
2
 g L
2  y  and
r 
T
T0
L  3y
2

2
L  2 yL
2
.
b) From the expression found in part a), T  T0 when y  23 L . At this point, a simple
pendulum with length y would have the same period as the meter stick without the
added mass; the two bodies oscillate with the same period and do not affect the other’s
motion.
13.100: Let the two distances from the center of mass be d 1 and d 2 . There are then two
relations of the form of Eq. (13.39); with I 1  I cm  md 12 and I 2  I cm  md 22 , these
relations may be rewritten as
mgd 1T
2
mgd 2 T
2
 4π
2
 4π
2
I
I
cm
 md 1
cm
 md
2
2
2

.
Subtracting the expressions gives
mg d 1  d 2 T
2
 4 π m d 1  d 2   4 π m  d 1  d 2  d 1  d 2 ,
2
2
2
2
and dividing by the common factor of m  d 1  d 2  and letting d 1  d 2  L gives the
desired result.
13.101: a) The spring, when stretched, provides an inward force; using ω  2 l for the
magnitude of the inward radial acceleration,
m ω  l  k l  l 0 , or l 
kl 0
k  m ω
2
.
b) The spring will tend to become unboundedly long.
13.102: Let r  R 0  x , so that r  R 0  x and
F  A[ e
 2 bx
e
 bx
].
1
When x is small compared to b , expanding the exponential function gives
F  A 1  2 bx   1  bx    Abx ,
corresponding to a force constant of Ab  579 . 2 N m or 579 N m to three figures. This is
close to the value given in Exercise 13.40.