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87. (a)
From the frequency information, we find
ω
= 2
π
f
= 10
π
rad/s.
A point on the
rope undergoing simple harmonic motion (discussed in Chapter 15) has maximum speed
as it passes through its "middle" point, which is equal to
y
m
ω
.
Thus,
5.0 m/s =
y
m
¡
y
m
= 0.16 m
.
(b) Because of the oscillation being in the
fundamental
mode (as illustrated in Fig. 16
23(a) in the textbook), we have
λ
= 2
L
= 4.0 m.
Therefore, the speed of waves along the
rope is
v
=
f
λ
= 20 m/s.
Then, with
μ
=
m
/
L
= 0.60 kg/m, Eq. 1626 leads to
v
=
τ
μ
¡
τ
=
v
2
= 240 N
2
2.4 10 N
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This note was uploaded on 04/16/2011 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
 Spring '09
 NAJAFZADEH
 mechanics, Simple Harmonic Motion

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