University Physics AI
No. 9 Waves
Class
Number
Name
I．Choose the Correct Answer
1. A disturbance can be written y ( x, t ) = (e
− ( x / b)2
e 2 xt / b e − t ) a . This disturbance is
2
(A) Not a traveling wave.
(B) A traveling wave with speed v = a.
(C) A traveling wave with speed v = a/b.
(D) A traveling wave with speed v = b.
Solution: y ( x, t ) = (e
−( x / b)2
e
2 xt / b − t 2 a
e
) =e
x
−( −t ) 2
b
=e
−
（
D ）
( x − bt ) 2
b2
= f ( x − vt )
Obviously, this disturbance is a traveling wave with speed v=b.
2. A traveling wave is of the form y ( x, t ) = A cos(kx − ωt ) + B sin(kx − ωt ) , which can also be
written as y ( x, t ) = D sin(kx − ωt − φ ) , where
(A) D = A + B
(B) D = A + B
（ C
(C) D = A + B
2
2
2
(D) D = A − B
3. Which of the following function is not a solution to the wave equation
(A) y = sin x cos t
2
∂2 y 1 ∂2 y
=
?（ D ）
∂x 2 v 2 ∂t 2
(B) y = tan( x + t )
(C) y = x − 6 x t + 12 xt − 8t
3
）
2
3
(D) y = sin( x + t ) cos( x − t )
Solution:
1
[sin( x + t ) + sin( x − t )] = f ( x + vt ) + f ( x − vt ) ,
2
(B) y = tan( x + t ) = f ( x + t ) ,
3
2
2
3
3
(C) y = x − 6 x t + 12 xt − 8t = ( x − 2t ) = f ( x − t ) ,
1
(D) y = sin( x + t ) cos( x − t ) = [sin 2 x − sin 2t ] ≠ f ( x ± t ) , it is not the solution of the wave
2
(A) y = sin x cos t =
equation.
4. Two waves travel down the same string. The waves have the same velocity, frequency (f0), and
wavelength but different phase constants (φ1>φ2) and amplitudes (A1>A2). According to the principle
of superposition, the resultant wave has a frequency f such that
(A) f = f 0 .
(B) f 0 / 2 < f < f 0
(C) 0 < f < f 0
（ A ）
(D) f = 2 f 0
Solution:
According to the principle of superposition, the resultant wave has a same frequency as any single
wave.
5. A standing wave occurs on a string when two waves of equal amplitude, frequency, and
wavelength move in opposite direction on a string. If the wavelength of two waves is decreased to
one-half the original length while the wave speed remains unchanged, then the angular frequency of
oscillation of the standing wave will
（
C
）
(A) decrease to one-half
. (B) remain the same
(C) double
Solution:
The wave length is λ = vT , if the v keeps unchanged, the wavelength decrease to one-half, then the
frequency of oscillation of the standing wave will double.
6. In a standing wave on a string, the spacing between nodes is ∆x . If the tension in the string is
doubled but the frequency of the standing waves is fixed, then the spacing between the nodes will
（ B
change to
(A) 2∆x
(B)
2∆x
）
(D) ∆x / 2
(C) ∆x / 2
Solution:
If the tension in the string is doubled, the velocity of the wave is v′ =
So the spacing between the nodes will change to ∆x′ =
λ′
2
=
v′f
=
2
T′
µ
=
2T
µ
= 2v
λ
2vf
= 2 = 2∆x
2
2
II. Filling the Blanks
1. A transverse wave on a string is described by the wavefunction
ψ ( x, t ) = (0.300 m) cos[(12.57 rad/m) x − (251.3 rad/s)t ]
The amplitude of the wave is 0.3m ; the angular wavenumber of the wave is 12.57rad/m ;
the wavelength of the wave is 0.5m ; the angular frequency of the wave is 251.3rad/s ; the
frequency of the wave is 40.0Hz ; the period of the wave is 0.025s ; the speed of the wave
.
is 20.0m/s ; the wave disturbance at x = 3.00 m when t = 1.50 × 10-3 s is 0.24m
Solution:
(a) A=0.3(m)
(b) k=12.57(rad/m)
(c) k =
2π
λ
⇒λ =
(d)
ω = 251.3rad/s
(e)
ν=
(f) T =
(g) v =
2π
2π
=
= 0.5(m)
12.57
k
ω 251.3
=
= 40.0(Hz)
2π
2π
2π
ω
ω
k
=
=
2π
= 0.025(s)
251.3
251.3
= 20.0(m/s)
12.57
−3
(h) Ψ = 0.3 cos(12.57 × 3 − 251.3 × 1.5 × 10 ) = 0.24(m)
2. A simple harmonic transverse wave is propagating
along a string toward the left (or -x) direction. Fig. 1
shows a plot of the displacement as a function of
position at time t=0. The string tension is 3.6N and its
linear density is 25g/m. The wave amplitude is 5m ,
the wavelength is
20m , the wave speed is
0.14s
, and the
144m/s , the wave period is
,
maximum speed of a particle in the string is
write an equation describing the traveling wave
Ψ ( x, t ) = 5 cos[2π (
y(m)
6
4
2
0
-2
-4
-6
x(m)
10
30
50
70
Fig.1
x
1
+ 12t ) − π ] .
40
5
Solution:
See the figure 1, we know the wave amplitude is 5m, and the wavelength is 40m.
T
The wave speed is v =
µ
=
3.6
= 12 m/s .
25 × 10 −3
φ
λ
40
The wave period is T = =
= 3.33 s
v 12
The phase constant of point O is -0.2π according to the reference circle.
According to the equation Ψ ( x, t ) = A cos[2π (
Ψ ( x, t ) = 5 cos[2π (
x
λ
+ vt ) + φ ] , the wave function is
x
1
+ 12t ) − π ]
40
5
3. A traverse wave pulse on a string has the wavefunction
ψ ( x, t ) =
The speed of the pulse is
Solution:
5 m/s
.
Since the classical wave equation is
∂Ψ
0.5( x − 5t )
=−
∂x
[2 + ( x − 5t ) 2 ]2
⇒
0.250 m 3
.
2.00 m 2 + [ x − (5.00 m/s)t ] 2
∂ 2Ψ 1 ∂ 2Ψ
−
=0
∂x 2 v 2 ∂t 2
∂ 2Ψ
2( x − 5t ) 2
0 .5
=
−
2
2 3
∂x
[2 + ( x − 5t ) ] [2 + ( x − 5t ) 2 ]2
90
∂Ψ
2.5( x − 5t )
=
∂t [2 + ( x − 5t ) 2 ]2
∂ 2Ψ
50( x − 5t ) 2
12.5
=
−
2
2 3
∂t
[2 + ( x − 5t ) ] [2 + ( x − 5t ) 2 ]2
⇒
∂ 2Ψ
∂t 2 = 5(m/s)
∂ 2Ψ
∂x 2
The speed of the pulse is v =
4. Figure 2 depicts the waveform of a traveling sinusoidal wave at two instants, when t = 0 s and
0.25 s later. The wave is traveling toward increasing values of x. ψ(x, t) =
0.5 sin(πx − π (8n + 6)t +
π
2
ψ(x)(m)
) .
when t=0s
Solution:
From the figure, we can know
0.4
λ = 2 m , A = 0 .5 m
0.2
The distance the wave traveling from t=0s to t=0.25s is
nλ + 1.5m n = 0,1,2 L
0
1.0
2.0
3.0
x(m
-0.2
-0.2
therefore
λ
Fig.2
n λ + 1.5m
T
0.25
0.25λ
1
⇒T =
=
n λ + 1.5m 4 n + 3
2π
⇒ω =
= π (8n + 6)
T
For k =
when t=0.250s
2π
λ
=
= π (rad/m)
Then we have
Ψ = A cos(kx − ωt ) = A sin(kx − ωt +
π
2
) = 0.5 sin(πx − π (8n + 6)t +
π
2
)
III. Give the Solutions of the Following Problems
1. A periodic transverse wave moving at 4.00 m/s toward
decreasing values of x has the waveform indicated in Figure 3
when t = 0 s.
(a) What is the wavelength? Indicate the wavelength on the
ψ(x, 0s)(m)
λ
0.250
x(m)
-3
0
3
Fig.3
6
waveform.
(b) Draw a graph of the disturbance at x = 0 m as a function of time. Indicate the period of the wave
on this graph. What is the period in seconds?
(c) Use the results of part (b) to determine the frequency and confirm that the speed is the product of
the frequency and wavelength as Equation 12.10 states.
(d) What is the maximum transverse speed?
Solution:
(a) The wavelength is 3m.
(b) from the figure, the period in seconds is T = 0.75(s) .
Ψ(m)
v←
0.25
-0.5
(c) The frequency is
As
ν=
T
0
1.5
t(s)
1
1
4
=
= (Hz)
T 0.75 3
4
3
ν ⋅ λ = ⋅ 3 = 4(m/s)
Thus the speed is the product of the frequency and wavelength: v = ν ⋅ λ .
(d) The maximum transverse speed is
vtrans = (
dΨ (0, t )
0.25
) max = maximum slop of the graph =
= 1(m/s)
dt
0.25
2. When t = 0 s, a pulse on a string is described by the equation
ψ ( x,0 s) =
1.00 m 3
2.50 m 2 + x 2
(a) Sketch ψ(x, 0s) as a function of x over the domain -10.00 m ≤ x < 10.00 m.
(b) The pulse moves parallel to − iˆ with a speed of 8.00 m/s. Find ψ(x, t).
Solution:
(a) The graph is shown in figure.
0.4
0.3
(b) The wave function is
0.2
0.1
-10
-5
5
10
Ψ ( x, t ) =
1
2.5 + ( x + 8t ) 2
3. A policeman is chasing a speeder along a straight interstate highway. Both the speeder and
trooper are moving at 150.0 km/h. The sire on the police cruiser has a frequency of 2.0 × 10 Hz.
3
What frequency is detected: (a) by the policeman? (b) by the speeder? (c) by a doomed, dumb
possum (Didelphis maarsupialis) watching the cars approach from a vantage point in the middle of
the road?
Solution:
(a) The frequency is detected by the policeman is
V
V
ν ′ = ν ( ) = ν = 2 × 10 3 (Hz)
(b) The frequency is detected by the speeder is
ν ′ =ν (
(c)
ν ′ =ν (
V
) =ν (
V − Vsouce
V − Vspeeder
V − Vsouce
) =ν (
343
343 − 15 ⋅
3
10
3600
V −0
) = ν = 2 × 10 3 (Hz)
V −0
) = 2.28 × 10 3 (Hz)
4. A wave is described by the wavefunction
ψ ( x, t ) = (2.00 × 10−2 m) cos[(180 rad/m) x − (24.0 rad/s)t ]
(a) What wave, when added to the given wave, will form a standing wave pattern?
(b) For such a standing wave, what is the distance between the nodes of the wave?
(c) For such a standing wave pattern, what is the distance between the antinodes?
(d) What is the distance between a node and the nearest antinodes?
Solution:
(a) The wavefunction is
ψ ( x, t ) = (2.00 × 10−2 m) cos[(180 rad/m) x + (24.0 rad/s)t ] .
(b) The distance between the nodes is
λ
2
=
1 2π π 3.14
×
= =
= 0.017 m
2 k
k 180
(c) The distance between the antinodes is the same as the nodes.
(d) The distance between a node and the nearest antinodes is
(n + 1)
λ
4
−n
λ
4
=
λ
4
= 0.0087 m