PHYS 2303 Homework 3 solutions
February 2017
1
Chapter 16
16.44 Distance moved by sound waves before hitting the cliff
∆x = vs
∆t
= 343 m/s
2
The girl can walk 1 km in 10 minutes.
v=
1 × 1000
= 1.67 m/s
10 × 60
t=
343
= 3.42 min
1.67
16.69 We know that the speed
p of waves on a string with tension T and linear
mass density µ is v = T /µ. Then,
r
600
vw1 =
= 489.9 m/s
0.0025
and,
r
600
= 414.04 m/s
0.0035
The pulses meet when the same amount of time has passed. If the pulses
meet at distance d, say from the left end,
vw2 =
t=
d
2−d
=
vw1
vw2
If we do some algebra, we arrive at
t=
2
= 2.21 ms
vw1 + vw2
16.73 We are given y(x, t) = A sin(6x − 24t). Reading off from the argument,
k = 6 m−1 and ω = 24 s−1
1
(a)
∂y
= −24A cos(6x − 24t)
∂t
We can identify the fact that vy,max = −24A. Hence |A| = 0.000125 m
vy =
(b) Using the expression for the wave velocity on a string, T = µv 2 .
Also, ω = vk. Using these
ω 2
T =µ
k
2
24
= 0.06 ×
6
= 0.96 N
16.88 If we consider the speaker as a point source, the intensity will fall off over
a sphere as r2 . Hence,
P
I=
4πr2
r
P
r=
4πI
r
120 × 10
=
4π × 3.82
= 5m
16.101 We have two waves with a relative phase difference: y1 = A cos(kx − ωt)
and y2 = A cos(kx − ωt + ϕ).
Y = A [cos(kx − ωt) + cos(kx − ωt + ϕ)]
ϕ
ϕ
= 2A cos −
cos kx − ωt +
2
2
ϕ
ϕ
= 2A cos
cos kx − ωt +
2
2
Cosines are even functions, meaning cos(−θ) = cos θ.
16.104 (a) For a standing wave with n nodes,
λn =
2L
n
1
2
= m
6
3
The tension on the string is produced by the hanging mass
r
mg
v=
= 57.15 m/s
µ
λ6 =
2
Then, f = v/λ = 171.46 Hz
(b) λ = v/f = 343/171.46 = 2 m
16.106 The first two normal modes are shown in the diagram. Since the bar is
mounted on the centre, there will be modes at that point.
λ1 = 2L
f1 =
v
2L
2
L
3
3v
f3 =
2L
λ3 =
Figure 1: First two normal modes
16.139 (a) The tension T in the string is balanced out by the hanging weight:
T = mg. The component of tension along the incline keeps the block from
sliding down.
T = 20g sin 45
= 20 × 9.8 × sin 45
= 138.57 N
From this, we can calculate m = 14.14 kg
(b)
r
v=
138.57
= 67.96 m/s
0.025
16.146 We need to first determine the tension in the two segments of string. Using
staticsFB sin 35 = 5g
5 × 9.8
FB =
= 85 N
sin 35
FA = FB cos 35
= 85 × cos 35
= 70 N
3
Evaluation of the velocities follow from the usual formula.
r
r
70
85
vA =
= 91 m/s vB =
= 100 m/s
0.0085
0.0085
2
Chapter 17
17.109 There are two pieces of information given:
fave =
f1 + f2
= 4100 Hz
2
and
f1 − f2 = 0.5 Hz
The second equation is due to the beat frequency and assumes that f1 >
f2 . To calculate the individual frequencies, the equations need to be solved
simultaneously.
0.5 + f2 + f2
= 4100
2
0.5 + 2f2 = 8200
f2 = 4099.75 Hz
And, f1 = 4099.75 + 0.5 = 4100.25 Hz
17.117 For this problem, we need to consider the most general equation for
Doppler effect of sound, in which which both the observer and the source
are moving. The observed frequency is given by
fo =
v ± vo
fs
v ∓ vs
Consider the case where bird A is the one flying at 15 m/s and the second
one is bird B. From A’s point of view, B is approaching and hence it would
hear the frequency go up. The sign convention in the equation must be
adjusted accordingly. Then,
fo,A =
330 + 15
× 3800 = 4229 Hz
330 − 20
Similarly, the same reasoning applies for B as from its point of view, A is
approaching.
330 + 20
fo,B =
× 3200 = 3556 Hz
330 − 15
17.132 The variation of speed of sound in air with temperature is given by:
r
T
vs = 331 1 +
273
4
When T = 22 ◦ C, vs = 344 m/s The equation that we have to use is
fo =
v
fs
v − vs
fs
= 29.8 m/s = 66 mph
vs = v 1 −
fo
17.138 For any wind instrument with one open end, λn = 2L/n. The fundamental
frequency occurs for n = 1.
v
2L
v
L=
2f1
= 0.29 m/s
f1 =
17.141 In the phenomenon of wave-interference, constructive and destructive interference occurs when the path-difference between two light rays equals
some integer or half-integer multiple of the wavelength. For maxima (constructive interference)
|r1 − r2 | = d sin θ = nλ
nλ
θ = arcsin
n = 0, ±1, ±2, ...
d
For minima (destructive interference)
1
d sin θ = n +
λ
2
1 λ
n = 0, ±1, ±2, ...
θ = arcsin n +
2 d
17.147 (a)
µ=
0.0009 kg
= 0.00391 kg/m
0.23 m
The velocity is
s
v=
T
=
µ
r
850
= 466.25 m/s
0.00391
(b) For the ninth mode,
λ9 =
2L
= 2 × 0.239 = 0.0511 m
9
5
(c)
f9 =
v
= unit[9142]Hz
λ
(d) The sound waves produced will also wave the same frequency
(e) At T = 24 ◦ 24,
r
vs = 331 1 +
24
= 345.2 m/s
273
v
f
345.2
=
9142
= 0.0378 m
λ=
6