Physics 295 Chapter 15 Solutions to Homework Problems
P15.3 x   4.00 m  cos 3.00 t    Compare this with x  A cost    to find
(a)
  2 f  3.00
or
P15.5
f  1.50 Hz
T
1
 0.667 s
f
(b)
A  4.00 m
(c)
   rad
(d)
x  t  0.250 s   4.00 m  cos1.75   2.83 m
(a)
At t  0 , x  0 and v is positive (to the right). Therefore, this situation
corresponds to x  A sin  t
and
v  vi cost
Since f  1.50 Hz ,
  2 f  3.00
Also, A  2.00 cm , so that
x   2.00 cm  sin 3.00 t
(b)
vmax  vi  A  2.00 3.00   6.00 cm s  18.8 cm s
The particle has this speed at t  0 and next at
(c)
T
1

s
2
3
amax  A 2  2.00 3.00   18.0 2 cm s2  178 cm s2
2
This positive value of acceleration first occurs at
(d)
t
3
t  T  0.500 s
4
2
s and A  2.00 cm , the particle will travel 8.00 cm in this time.
3
 3 
Hence, in 1.00 s  T  , the particle will travel
 2 
Since T 
8.00 cm  4.00 cm  12.0 cm .
P15.10 m  1.00 kg , k  25.0 N m , and A  3.00 cm . At t  0 , x  3.00 cm
(a)
k
25.0

 5.00 rad s
m
1.00
2
2
so that,
T

 1.26 s
 5.00
(b)
vmax  A  3.00 102 m  5.00 rad s  0.150 m s

amax  A 2  3.00  102 m  5.00 rad s  0.750 m s2
2
(c)
Because x  3.00 cm and v  0 at t  0 , the required solution is x   A cos t
or
x  3.00cos 5.00t  cm
dx
 15.0sin  5.00t  cm s
dt
dv
a
 75.0cos 5.00t  cm s2
dt
v
P15.15 Choose the car with its shock-absorbing bumper as the system; by conservation of
energy,
k
5.00  106
1 2 1 2
  3.16  102 m 
 2.23 m s
mv  kx : v  x
m
103
2
2
P15.19 Model the oscillator as a block-spring system.
From energy considerations,
v2   2 x2   2 A 2
vmax   A and v 
A
2
 A 
2 2
2 2

  x   A
 2 
2
so
3
4
From this we find x2  A 2 and
x
P15.24
3
A  2.60 cm
2
where A  3.00 cm
The period in Tokyo is
TT  2
LT
gT
and the period in Cambridge is TC  2
We know TT  TC  2.00 s
For which, we see
LT LC

gT gC
LC
gC
gC LC 0.994 2


 1.001 5
gT LT 0.992 7
or
P15.25
Using the simple harmonic motion model:
A  r  1 m 15


180
 0.262 m
g
9.8 m s2

 3.13 rad s
L
1m
(a)
vmax  A  0.262 m 3.13 s  0.820 m s
(b)
amax  A 2  0.262 m  3.13 s  2.57 m s2
2
atan  r
(c)

atan 2.57 m s2

 2.57 rad s2
r
1m
FIG. P15.25
F  ma  0.25 kg 2.57 m s2  0.641 N
More precisely,
(a)
mgh 
 vmax
(b)
I  mgL sin 
 max 
(c)
1 2
and
h  L 1  cos 
mv
2
 2gL 1  cos   0.817 m s
mgL sin  g
 sin i  2.54 rad s2
mL2
L
Fmax  mg sin i  0.250 9.80sin 15.0  0.634 N
The answers agree to two digits. The answers computed from conservation of energy and
from Newton’s second law are more precisely correct. With this amplitude the motion of
the pendulum is approximately simple harmonic.