Chapter 17 – Student Solutions Manual
5. Let
t
f
be the time for the stone to fall to the water and
t
s
be the time for the sound of the
splash to travel from the water to the top of the well. Then, the total time elapsed from
dropping the stone to hearing the splash is
t
=
t
f
+
t
s
. If
d
is the depth of the well, then the
kinematics of free fall gives
2
1
2
f
d
gt
=
, or
2
/
.
f
t
d
=
g
The sound travels at a constant
speed
v
s
, so
d
=
v
s
t
s
, or
t
s
=
d
/
v
s
. Thus the total time is
2
/
/
s
t
d
g
d
=
+
v
. This equation is
to be solved for
d
. Rewrite it as
2
/
/
s
d
g
t
d
v
=
−
and square both sides to obtain
2
d
/
g
=
t
2
– 2(
t
/
v
s
)
d
+ (1 +
2
s
v
)
d
2
.
Now multiply by
g
2
s
v
and rearrange to get
gd
2
– 2
v
s
(
gt
+
v
s
)
d
+
g
2
s
v
t
2
= 0.
This is a quadratic equation for
d
. Its solutions are
(
)
2
2
2
2
(
)
4
4
.
2
s
s
s
s
s
v
gt
v
v
gt
v
g v t
d
g
+
±
+
−
=
2
2
The physical solution must yield
d
= 0 for
t
= 0, so we take the solution with the negative
sign in front of the square root. Once values are substituted the result
d
= 40.7 m is
obtained.
7. If
d
is the distance from the location of the earthquake to the seismograph and
v
s
is the
speed of the S waves then the time for these waves to reach the seismograph is
t
s
. =
d
/
v
s
.
Similarly, the time for P waves to reach the seismograph is
t
p
=
d
/
v
p
. The time delay is
Δ
t
= (
d
/
v
s
) – (
d
/
v
p
) =
d
(
v
p
–
v
s
)/
v
s
v
p
,
so
3
(4.5 km/s)(8.0km/s)(3.0min)(60s /min)
1.9
10 km.
(
)
8.0km/s
4.5km/s
s
p
p
s
v v
t
d
v
v
Δ
=
=
=
×
−
−
We note that values for the speeds were substituted as given, in km/s, but that the value
for the time delay was converted from minutes to seconds.
9. (a) Using
λ
=
v
/
f
, where
v
is the speed of sound in air and
f
is the frequency, we find
5
6
343m/s
7.62
10
m.
4.50
10 Hz
−
λ =
=
×
×

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