Phys 2214 Homework #4 Solutions
September 30, 2011
1. For a long, thin rod, the compressions created by a longitudinal wave
traveling along the rod will change the length to a greater extent than
the diameter. So, we can replace the equation for the bulk modulus,
p
=

B
(Δ
V/V
), by the equation for Young’s modulus,
F/A
=
Y
(Δ
x/x
).
s(x+Δx,t)
x+Δx
x
s(x,t)
(a) In general, the change in length of the segment is
Δ
L
=
[
x
+ Δ
x
+
s
(
x
+ Δ
x, t
)]

[
x
+
s
(
x, t
)]

Δ
x
=
s
(
x
+ Δ
x, t
) +
x
+ Δ
x

x

Δ
x

s
(
x, t
)
=
s
(
x
+ Δ
x, t
)

s
(
x, t
)
.
1
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(b) The linear stressstrain relation is
F
A
=
Y
Δ
L
L
=
Y
s
(
x, t
)

s
(
x
+ Δ
x, t
)
Δ
x
,
where the fact that Δ
L
is negative has been used and the original
length is Δ
x
.
(c) The force is given by
F
(
x, t
)
=
lim
Δ
x
→
0
Y A
s
(
x, t
)

s
(
x
+ Δ
x, t
)
Δ
x
=

Y A
∂s
∂x
.
(d) The net force acting on the segment is
F
(
x, t
)

F
(
x
+ Δ
x, t
).
Then Newton’s 2
nd
law gives

Y A
"
∂s
(
x, t
)
∂x

∂s
(
x
+ Δ
x, t
)
∂x
#
=
ρ
0
A
Δ
x
∂
2
s
∂t
2
.
Dividing both sides by Δ
x
and taking the limit as Δ
x
→
0 gives
lim
Δ
x
→
0
h
∂s
(
x,t
)
∂x

∂s
(
x
+Δ
x,t
)
∂x
i
Δ
x
=

∂
2
s
∂x
2
.
We then get the wave equation
∂
2
s
∂x
2
=
ρ
0
Y
∂
2
s
∂t
2
,
which, upon comparison with the general wave equation, gives
v
=
q
Y/ρ
0
.
2. The fundamental frequency is given as
f
1
= 150
Hz
.
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 '11
 DAVIS,J.C.
 Work, Waves And Optics, Frequency, Hertz, Wavelength, Hz

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