Physics 305
1)
Homework Solutions
Waves and Wavefunctions
Page 1
The figure below represents the profile (t = 0 s) of a transverse wave on a string traveling in the
positive x–direction at a speed of 100 cm/s. (The values for x and y are in cm.)
20.0 y(x,0)
(cm)
10.0
-20.0
0.0
-10.0
0.0
10.0
20.0
30.0
-10.0
40.0
x (cm)
-20.0
(a) Determine its wavelength.
There is a trough in the wave at x = 0.0 cm and again at x = 30.0 cm, so the wave
repeats every 30 cm. Therefore, the wavelength is given by λ = 30 cm.
(b) Notice that as the wave passes any fixed point on the x–axis the string at that location
oscillates in time. Draw a graph of y versus t showing how a point on the rope at x = 0
oscillates.
Since the wave has a speed of 100 cm/s, the wave trough at x = 0.0 cm will take
0.3 seconds to reach x = 30.0 cm. Therefore, the fixed point at x = 0.0 cm will go
through one cycle in 0.3 s. The amplitude of the motion will be 20.0 cm, since it
looks like the wave has a maximum displacement of 20.0 cm. The wave is not
symmetric, so it will take 0.2 s to go from maximum negative displace to
maximum positive displacement, and then another 0.1 s to return to maximum
negative displacement.
20.0 y(0,t)
(cm)
10.0
-0.1
0.0
0.0
0.1
0.2
-10.0
0.3
t (s)
-20.0
(c) What is the frequency of the wave?
The frequency can be found two different ways:
f=
1cycle
= 3.33Hz
0.3s
or
f =
v 100 cm s
=
= 3.33Hz
λ
30 cm
Physics 305
2)
Homework Solutions
Waves and Wavefunctions
Page 2
A transverse wave on a string travels in the negative x–direction at a speed of 40.0 cm/s. The
figure below is a graph of y versus t showing how a point on the rope at x = 0 oscillates. (The
values for y are in cm.)
2.0 y(0,t)
(cm)
1.0
0.0
-0.1
0.0
0.1
0.2
0.3
t (s)
-1.0
-2.0
(a) Determine the wave’s period.
The motion appears to repeat every 0.2 s, which is the definition of period.
Therefore, the period is given by T = 0.2 s.
(b) What is the frequency of the wave?
The frequency is the reciprocal of the period, so:
f=
1
= 5.0 Hz
0.2 s
(c) What is the wavelength of the wave?
The wavelength is related the wave speed and the frequency by:
λ=
v 40 cm s
=
= 8.0 cm
f 5.0 Hz
(d) Sketch the profile of the wave (y versus x)
The motion appears to be symmetric, so we need to draw a symmetric, triangular
wave that has a wavelength of 8.0 cm and an amplitude of 2.0 cm.
2.0 y(x,0)
(cm)
1.0
-10.0
-5.0
0.0
-1.0
-2.0
0.0
5.0
10.0
15.0
20.0
x (cm)
25.0
Physics 305
3)
Homework Solutions
Waves and Wavefunctions
Page 3
The wavefunction of a transverse wave on a string is
'!
rad $ !
rad $ *
y ( x,t ) = (10.0 cm ) cos )# 942.5
& x + #15.0
&t ,
cm % "
s %+
("
Determine the (a) frequency, (b) wavelength, (c) period, (d) amplitude, (e) propagation constant,
(f) phase velocity, and (g) angular frequency.
Perhaps the simplest way to approach this problem is to write the wavefunction in
a couple of its different general forms, and then just pick out the matching parts.
y ( x,t ) = A cos [ k x – ω t ]
% 2π
2π (
= A cos '
x − t*
&λ
T )
= A cos [ 2 πκ x −2 π f t ]
= A cos %&k ( x − vt )()
where A is the amplitude, k is the propagation constant, λ is the wavelength, λ is
the wavelength, T is the period, κ is the wave number, f is the frequency, and v is
the wave speed.
So, comparing the given wavefunction with the general forms, we can see that:
a) 2 π f = 15.0
rad
s
so, f =
15.0 rad s 7.5
=
Hz = 2.39 Hz
2π rad
π
b)
2π
rad
= 942.5
λ
cm
so, λ =
c)
2π
rad
= 15.0
T
s
so, T =
2π rad
= 6.67 × 10 −3 cm
942.5 rad cm
2π rad
π s = 0.419s
=
15.0 rad s 7.5
d) A =10.0cm
e) k = 942.5
f)
rad
cm
ω
15 rad/s
cm
. The sign in front of the ω t term is
=
= 1.59 × 10 −2
k 942.5 rad/cm
s
cm
!
positive, so the velocity is v = –1.59 × 10 −2
x̂
s
vp =
g) ω = 15.0
rad
s
Physics 305
4)
Homework Solutions
Waves and Wavefunctions
Page 4
Does the following function, in which C is a constant,
" x%
z ( x,t ) = C $t − '
# v&
represent a wave?
The argument can certainly be arranged to be in the form (x – v t) but the function
is not twice differentiable, so this does not represent a wave.
5)
Determine which of the following describe traveling waves:
1)
y ( x, t ) = e
(
– a 2 x 2 +b 2 t 2 −2abtx
(
)
)
2)
y ( z, t ) = A sin az 2 – bt 2
3)
⎛x t⎞
y ( x, t ) = A sin 2π ⎜ + ⎟
⎝ a b⎠
4)
y ( x, t ) = A cos 2 2π ( t – x )
2
Where appropriate, draw the profile and find the speed and direction of motion.
A wavefunction that describes a traveling wave needs to be twice differentiable
and the argument needs to be in the form ( x ∓ vt ) where v is the wave speed, a
minus sign represents a wave traveling in the positive x-direction, and a plus sign
represents a wave traveling in the negative x-direction.
1) The exponential function is twice differentiable. The argument of the function
is − a 2 x 2 + b 2t 2 − 2abt x , so
(
)
− ( a x + b t − 2abtx ) = − ( ax − bt )
2
2
2 2
2
2
b ⎞
b
!
⎛
= −a ⎜ x − t ⎟ ⇒ v = + x̂
⎝
a ⎠
a
2
1.2
1.0
0.8
0.6
0.4
0.2
!20
!10
0
a = 0.1
10
1
m
b = 1.0
20
1
t = 1.5 s
s
30
40
Physics 305
Homework Solutions
Waves and Wavefunctions
Page 5
2)
The sine function is twice differentiable. The argument of the function is
2
az − bt 2 , so
(
( az
2
− bt 2
)
)
=
(
az + bt
)(
az − bt
)
(
)
This cannot be rearranged into the form c1 x  c2 t , so this function does not
represent a traveling wave.
3) The sine function is twice differentiable. The argument of the function is
2
⎛x t⎞
2π ⎜ + ⎟ , so
⎝ a b⎠
⎛x t⎞
2π ⎜ + ⎟
⎝ a b⎠
2
⎛ x at ⎞
= 2π ⎜ + ⎟
⎝ a ab ⎠
2π ⎛
a ⎞
= 2 ⎜ x + t⎟
a ⎝
b ⎠
2
2
a
!
⇒ v = − x̂
b
1.0
0.5
!3
!2
!1
1
!0.5
!1.0
a = 1.0 m b = 1.0 s t = 0.5 s
2
3
Physics 305
Homework Solutions
Waves and Wavefunctions
Page 6
4) The cosine function is twice differentiable. The argument of the function is
2π ( t − x ) , so
2π ( t − x ) = −2π ( x − t )
(since cos
= 2π ( x − t )
2
is an even function
)
m

⇒ v = +1 x̂
s
1.2
1.0
0.8
0.6
0.4
0.2
!3
!2
!1
0
1
2
3
a = 1.0 m b = 1.0 s t = 0.5 s
6)
2
A pulse of the form y = ae–bx is formed in a rope, where a and b are constants and x is in
centimeters. Sketch this pulse. Then write an equation that represents the pulse moving in the
negative direction at 10 cm/s.
2.0
1.5
1.0
0.5
!20
!15
!10
!5
0
a = 1.8 cm b = 0.1
5
1
cm 2
10
Physics 305
Homework Solutions
Waves and Wavefunctions
Page 7
To make this pulse into a traveling wave pulse moving in the negative direction,
we just need to replace the argument “x” with “ x + vt ”, where v is 10 cm/s.
So, y ( x,t ) = ae
7)
cm ⎞
⎛
−b⎜ x+10 t ⎟
⎝
s ⎠
2
A transverse wave pulse, described by
y ( x,0 ) =
4
x2 + 2
is initiated at t = 0 in a stretched string.
(a) I seem to have forgotten to include the units with the two numbers (bad, bad me). What units
does the 4 have, and what units does the 2 have?
y ( x,0 ) represents a displacement, so it must be in units of meters, [m]. x
represents position [m] and is squared, so the value 2 must have the units of [m]2.
Therefore, the value of 4 must have the units of [m]3.
(b) Write an equation for the traveling pulse if it moves with a speed of 2.5 m/s in the negative xdirection.
To make this pulse into a traveling wave pulse moving in the negative direction,
we just need to replace the argument “x” with “ x + vt ”, where v is 2.5 m/s.
So, y ( x,t ) =
4 m3
2
m ⎞
⎛
2
⎜⎝ x + 2.5 t ⎟⎠ + 2 m
s
(c) Plot (or graph) the pulse at t = 0, t = 2, and t = 5 seconds,
Physics 305
8)
Homework Solutions
Waves and Wavefunctions
Page 8
Consider the following mathematical expressions, where distances are in meters:
1)
2)
3)
y ( z, t ) = A sin 2 ⎡⎣ 4π ( t + z ) ⎤⎦
y ( x, t ) = A ( x – t )
(
2
y ( x, t ) = A Bx 2 – t
)
(a) Which qualify as traveling waves? Justify your conclusion.
Similar to problem 4, if we can rearrange the argument of any of the above
functions to look like x ∓ vt , and the function is twice differentiable, then that
function represents a traveling wave with wave speed v. Expressions 1) and 2)
satisfy these criteria, but expression 3) does not. Therefore, functions 1) and 2)
qualify as traveling waves.
(b) If they qualify, give the magnitude and direction of the wave velocity.
1) The argument of the function is 4π ( t + z ) , so
m

4π ( t + z ) = 4π ( z + t ) ⇒ v = −1 ẑ
s
m

2
2) The argument of the function is ( x − t ) , so v = 1 x̂
s
9)
A harmonic traveling wave is moving in the negative z–direction with an amplitude of 2 m, a
wavelength of 5 m, and a period of 3 s. Its displacement at the origin is zero at time zero. Write a
wavefunction for this traveling wave
As with problem 4, the simplest way to approach this problem is to write the
wavefunction in a couple of its different general forms, and then just substitute in
the appropriate values. The wave is harmonic, so we need a sin or cos function.
Since the displacement at the origin is zero at time zero, it would be easier to use
a sin function instead of a cos function, so that we don’t have to include an initial
phase. We are using a plus sign since the wave is moving in the negative z–
direction.
y ( z,t ) = Asin [ kz + ω t ]
2π ⎤
⎡ 2π
= Asin ⎢ z + t ⎥
T ⎦
⎣λ
= Asin [ 2πκ z +2π f t ]
= Asin ⎡⎣ k ( z + vt ) ⎤⎦
Physics 305
Homework Solutions
Waves and Wavefunctions
Page 9
(a) that exhibits directly both wavelength and period;
Using the second form, with A = 2 m, λ = 5 m, and T = 3 s, we can write:
⎡ ⎛ z t ⎞⎤
y ( z,t ) = Asin ⎢ 2π ⎜ + ⎟ ⎥
⎣ ⎝ λ T ⎠⎦
⎡ ⎛ z
t ⎞⎤
= ( 2m ) sin ⎢ 2π ⎜
+ ⎟⎥
⎣ ⎝ 5m 3s ⎠ ⎦
(b) that exhibits directly both propagation constant and velocity;
The propagation constant is given by k ≡
v= λf =
2π
and the velocity is given by
λ
λ
. Using the fourth form, with A = 2 m, λ = 5 m, and T = 3 s, we can
T
write:
y ( z,t ) = Asin ⎡⎣ k ( z + vt ) ⎤⎦
⎡ 2π ⎛
5m ⎞ ⎤
= ( 2m ) sin ⎢
z
+
t ⎥
⎜
3s ⎟⎠ ⎦
⎣5m⎝
(c) in complex form.
Using k ≡
2π
2π
and ω =
, with λ = 5 m, and T = 3 s, we can write:
λ
T
y ( z,t ) = Asin ( kz + ω t )
{
= A Im ⎡⎣ cos ( kz + ω t ) + i sin ( kz + ω t ) ⎤⎦
{
= A Im ⎡⎣ ei ( kz + ω t ) ⎤⎦
}
⎧⎪ ⎡ i ⎛⎜⎝ 2 π z + 2 π t ⎞⎟⎠ ⎤ ⎫⎪
= ( 2m ) ⎨Im ⎢ e 5m 3s ⎥ ⎬
⎥⎦ ⎪⎭
⎪⎩ ⎢⎣
}
Physics 305
10)
Homework Solutions
Waves and Wavefunctions
Page 10
Two waves of the same amplitude, speed, and frequency travel together in the same region of
space. The resultant wave may be written as a sum of the individual waves,
y ( x, t ) = A sin ( kx + ω t ) + A sin ( kx – ω t + π )
With the help of complex exponentials, show that
y ( x, t ) = 2A cos ( kx ) sin (ω t )
y ( x,t ) = Asin ( kx + ω t ) + Asin ( kx – ω t + π )
{
= A Im ⎡⎣ cos ( kx + ω t ) + i sin ( kx + ω t ) ⎤⎦ + Im ⎡⎣ cos ( kx – ω t + π ) + i sin ( kx – ω t + π ) ⎤⎦
{
= A {Im ⎡⎣ e
= A Im ⎡⎣ ei( kx+ω t ) ⎤⎦ + Im ⎡⎣ ei( kx–ω t+π ) ⎤⎦
{
= A {Im ⎡⎣ e
i( kx+ω t )
+ ei( kx–ω t+π ) ⎤⎦
}
= A Im ⎡⎣ eikx eiω t + eikx e–iω t eiπ ⎤⎦
{
ikx iω t
e
– eikx e–iω t ⎤⎦
= A Im ⎡⎣ eikx ( eiω t – e–iω t ) ⎤⎦
}
}
} (since e
}
iπ
= –1)
⎧⎪ ⎡
⎛ eiω t – e–iω t ⎞ ⎤ ⎫⎪ ⎛
⎞
eiω t – e–iω t
= A ⎨Im ⎢ 2ieikx ⎜
since
= sin (ω t )⎟
⎬
⎥ ⎜
⎟
2i
2i
⎝
⎠ ⎦ ⎭⎪ ⎝
⎠
⎩⎪ ⎣
{
= 2A Im ⎡⎣ieikx sin (ω t ) ⎤⎦
{
}
}
= 2A Im ⎡⎣i ( cos ( kx ) + i sin ( kx )) ⎤⎦ sin (ω t )
{
}
= 2A Im ⎡⎣ – sin ( kx ) + i cos ( kx ) ⎤⎦ sin (ω t )
= 2A cos ( kx ) sin (ω t )
}