Answers Chapter 11
3.The spring constant is the ratio of external applied force to displacement.
k=
4.
The period is 2.0 seconds, and the mass is 32 kg. The spring constant is calculated from Eq. 11–6a.
T = 2p
6.
8.
Fext
210 N - 75 N
135 N
=
=
= 562.5 N/m » 560 N/m
x
0.85 m - 0.61 m 0.24 m
m
m
m
32 kg
® T 2 = 4p 2
® k = 4p 2 2 = 4p 2
= 315.8 N/m » 320 N/m
k
k
T
(2.0 s)2
The relationship between frequency, mass, and spring constant is Eq. 11–6b, f =
(a)
f =
1
2p
k
m
(b)
f =
1
2p
k
1
=
m 2p
1
2p
k
.
m
® k = 4p 2 f 2 m = 4p 2 (4.0 Hz)2 (2.2 ´10-4 kg) = 0.1390 N/m » 0.14 N/m
0.1390 N/m
4.4 ´10-4 kg
= 2.828 Hz » 2.8 Hz
We assume that downward is the positive direction of motion. For this motion, we have
k = 305 N/m, A = 0.280 m, m = 0.235 kg, and w = k/m = (305 N/m)/0.235 kg = 36.026 rad/s.
(a)
Since the mass has a zero displacement and a positive velocity at t = 0, the equation is a sine
function.
y(t) = (0.280 m) sin [(36.0 rad/s)t]
(b)
2p
= 0.17441 s. The spring will have
w 36.026 rad/s
its maximum extension at times given by the following:
The period of oscillation is given by T =
2p
=
T
+ nT = 4.36 ´10-2 s + n(0.174 s), n = 0,1,2,
4
The spring will have its minimum extension at times given by the following:
tmax =
tmin =
10.
3T
+ nT = 1.31´10-1 s + n(0.174 s), n = 0,1,2,
4
(a) We find the effective spring constant from the mass and the frequency of oscillation,
using
Eq. 11–6b.
f =
1
2p
k
m
®
k = 4p 2 mf 2 = 4p 2 (0.052 kg)(3.0 Hz)2 = 18.476 N/m » 18 N/m
(b)
Since the objects are the same size and shape, we anticipate that the spring constant is the
same.
f =
1
2p
k
1 18.476 N/m
=
= 1.293 Hz » 1.3 Hz
m 2p
0.28 kg
27.The period of a pendulum is given by T = 2p
pendulum both on Mars and on Earth.
T = 2p
/g ®
TMars = TEarth
31.
TMars 2p
=
TEarth 2p
/g. The length is assumed to be the same for the
/g Mars
/g Earth
=
g Earth
g Mars
®
g Earth
1
= (1.85 s)
= 3.0 s
g Mars
0.37
30.
The period of a pendulum is given by Eq. 11–11a, T = 2p
(a)
T = 2p
(b)
If the pendulum is in free fall, there is no tension in the string supporting the pendulum bob
and no restoring force to cause oscillations. Thus there will be no period—the pendulum will
not oscillate, so no period can be defined.
/g = 2p
0.47 m
9.80 m/s2
/g.
= 1.4 s
There are (24 h)(60 min/h)(60 s/min) = 86,400 s in a day. The clock should make one cycle in
exactly two seconds (a “tick” and a “tock”), so the clock should make 43,200 cycles per day. After
one day, the clock in question is 21 seconds slow, which means that it has made 10.5 fewer cycles
than are required for precise timekeeping. Thus the clock is only making 43,189.5 cycles in a day.
43,189.5
Accordingly, the period of the clock must be decreased by a factor of
.
43,200
Tnew =
new
43,189.5
T
® 2p
43,200 old
æ 43,189.5 ö
=ç
è 43,200 ÷ø
2
new /g
æ 43,189.5 ö
=ç
2p
è 43,200 ÷ø
old /g
®
2
old
æ 43,189.5 ö
=ç
(0.9930 m) = 0.9925 m
è 43,200 ÷ø
Thus the pendulum should be shortened by 0.5 mm.
33.
If we consider the pendulum as starting from its maximum displacement, then the equation of
2p t
. Solve for the time for the position to decrease
motion can be written as q = q0 cos w t = q 0 cos
T
to half of the amplitude.
q1/2 = 12 q0 = q0 cos
2p t1/2
2p t1/2
p
®
= cos-1 12 =
® t1/2 = 61 T
T
T
3