Chapter 13 Solutions - Mosinee School District

Chapter 13
L
9.80 m s2
2.00 s
gT
2
4 4
2 2
2
0.993 m
Then mass per unit length of the string is then
m
L
0.060 0 kg kg
0.060 4
0.993 m m
When the pendulum is vertical and stationary, the tension in the string is
F Mball
g
5.00 kg 9.80 m s2
49.0 N
and the speed of transverse waves in it is
v
F
49.0 N
0.060 4 kg m
28.5 m s
13.56 If
1 m1
L is the mass per unit length for the first string, then
2 m2 L m1 2L 1 2 is that of the
second string. Thus, with F2 F1
F , the speed of waves in the second string is
F 2 F F
v 2 2 v 2 5.00 m s 7.07 m s
2 1
2 1 1
13.57 (a) The tension in the string is F mg 3.0 kg 9.80 m s2
29 N . Then, from v F , the mass per
unit length is
F
v2
29 N
24 m s
2
0.051 kg m
(b)
When m = 2.00 kg, the tension is
F mg
2.0 kg 9.80 m s2
20 N
and the speed of transverse waves in the string is
v
F
20 N
0.051 kg m
20 m s
13.58 If the tension in the wire is F, the tensile stress is Stress = F/A, so the speed of transverse waves in the wire may be
written as
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