Physics 9 Fall 2009
Homework 12 - Solutions
1. Chapter 21 - Exercise 8.
The figure shows a standing wave that is oscillating at frequency f0 .
(a) How many antinodes will there be if the frequency is doubled to 2f0 ? Explain.
(b) If the tension in the string is increased by a
factor of four, for what frequency, in terms
of f0 , will the string continue to oscillate as
a standing wave with three antinodes?
————————————————————————————————————
Solution
(a) Because the ends of the string are tied down, increasing the oscillation frequency
will decrease the wavelength. Doubling the frequency will cut the wavelength in
half, and so will double the number of antinodes to six.
q
T
.
(b) The velocity of the wave depends on the square root of the tension, v =
µ
So, increasing the tension by a factor of 4 doubles the velocity. Since the velocity
of the wave is v = λf , if we double v, but want to keep the wavelength fixed
(this fixes the number of antinodes), then we would need to double the frequency,
which we can see as follows: v = λf0 → 2v = λ (2f0 ) = λf . So, the new frequency
f = 2f0 .
1
2. Chapter 21 - Exercise 18.
The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00
m long and has a mass of 400 g. The vibrating section of the string is 1.90 m long.
What tension is needed to tune this string properly?
————————————————————————————————————
Solution
The lowest note on the grand piano is the fundamental harmonic, meaning that the
wavelength is twice the length of the vibrating section of the string, λ = 2L, where
L = 1.90 m. q
The speed of the wave is v = λf , but also depends on the tension of the
wire as v = Tµ , where µ = m/` is the linear mass density of the wire (` is the full
length of the wire, not the vibrating part). So, solving for the tension gives
T = µ (λf )2 = 4µL2 f 2 = 4
mL2 f 2
.
`
With numbers we find
mL2 f 2
.400 × (1.90)2 (27.5)2
T =4
=4
= 2185 N,
`
2
which is a pretty big amount of force!
2
3. Chapter 21 - Exercise 19.
A violin string is 30 cm long. It sounds the musical note A (440 Hz) when played
without fingering. How far from the end of the string should you place your finger to
play the note C (523 Hz)?
————————————————————————————————————
Solution
Since the speed ofp
the wave along the string depends on the tension and mass density
of the string, v = T /µ, it is constant. Furthermore, the speed of the wave is related
to its frequency and wavelength, v = λf . So, λ440 f440 = λ523 f523 . Since the 440 Hz
note is the fundamental frequency, played without fingering, the wavelength λ440 =
2L440 = 2 × 30 = 60 cm. The wavelength λ523 = 2L523 , and so
L523 =
440
f440
L440 =
× 30 = 25.2 cm.
4523
523
So, the length of the string is 25.2 cm. But, the question asks how far from the edge
you should put your finger. Since the sting is 30 cm long, but you need it to be 25.2
cm, you should put your finger back 30 − 25.2 = 4.8 cm from the end of the string.
3
4. Chapter 21 - Exercise 24.
A very thin oil film (n=1.25) floats on water (n=1.33). What is the thinnest film that
produces a strong reflection for green light with a wavelength of 500 nm?
————————————————————————————————————
Solution
The strong reflection occurs when the light reflected from the oil and water interfere
constructively. The extra distance that the wave reflected from the water travels is 2d,
where d is the thickness of the oil (it has to go through the oil, then back again). This
distance has to be a whole number of wavelengths, mλ. But, in a material of index n,
the wavelength is decreased by a factor of n. Here n is the index of the foil, since this
. Since we want the
is the material through which the light is traveling. So, 2d = mλ
n
thinnest film, we’ll set m = 1. Then,
d=
500
λ
=
= 200 nm.
2n
2 (1.25)
4
5. Chapter 21 - Problem 41.
Astronauts visiting Planet X have a 2.5 m-long string whose mass is 5.0 g. They tie
the string to a support, stretch it horizontally over a pulley 2.0 m away, and hang a
1.0 kg mass on the free end. Then the astronauts begin to excite standing waves on
the string. Their data show that standing waves exist at frequencies of 64 Hz and 80
Hz, but at no frequencies in between. What is the value of g, the free-fall acceleration
on Planet X?
————————————————————————————————————
Solution
The setup is seen in the figure to the right. The
speed of theq
wave depends on the tension in the
T
, where µ is the mass density of
string, v =
µ
the string. The tension in the string comes from
the weight
of the 1 kg mass, so T = M g. So,
q
Mg
. Again, though, the velocity is v = λf .
v=
µ
We know two frequencies, but they don’t tell us
the wavelength for each.
Higher harmonic frequencies increase as fm = mf0 , where f0 is the fundamental frequency, for which λ0 = 2L. We know that we have two frequencies, one at 64 Hz,
and the next at 80 Hz. From this we can determine the fundamental frequency. Suppose that the 64 Hz frequency is the mth frequency, mf0 = 64. Then, the (m + 1)th
frequency is (m + 1) f0 = 80. Subtracting the first from the second gives
(m + 1) f0 − mf0 = f0 = 80 − 64 = 16.
Thus, the fundamental frequency is f0 = 16 Hz. So, putting it all together gives
s
Mg
λ2 f 2 m
v = λ0 f0 =
⇒g= 0 0
,
µ
` M
where we have substituted µ = m/`, where m is the mass of the string, and ` is the
full length of the string. Finally, the fundamental wavelength is λ0 = 2L, and so
4L2 f02 m
g=
.
` M
Inserting values gives
4L2 f02 m
4 (2)2 (16)2 0.005
g=
=
= 8.2 m/s2 ,
` M
2.5
1
which is a fair amount smaller than Earth’s gravity.
5
6. Chapter 21 - Problem 66.
A soap bubble is essentially a very thin film of water (n=1.33) surrounded by air. The
colors that you see in soap bubbles are produced by interference, much like the colors
of dichroic glass.
(a) Derive an expression for the wavelengths λC for which constructive interference
causes a strong reflection from a soap bubble of thickness d.
Hint: Think about the reflection phase at both boundaries.
(b) What visible wavelengths of light are strongly reflected from a 390-nm-thick soap
bubble? What color would such a soap bubble appear to be?
————————————————————————————————————
Solution
(a) The light ray falling on the soap bubble reflects from two places. It reflects from
the surface of the bubble, and then it also reflects back from the back of the
bubble, after traveling an extra distance ∆x = 2d (down and back), where d is
the thickness of the bubble. The wave reflecting from the top surface reflects from
a material having a higher index of refraction. This induces an extra phase shift
of π, that the reflection from the back surface doesn’t have.
The net phase of the first wave, at a fixed time which we take to be t = 0, is
Φ1 = kx1 + φ1 + π, where k is the wave number, and φ1 is the intrinsic phase shift
due to the source. Similarly, the net phase of the second wave isΦ2 = kx2 + φ2 .
The second wave doesn’t have an extra phase shift because it’s reflecting from the
air behind the bubble, which has a lower index of refraction. The phase shifts of
these two waves are equal since they both come from the same source, so φ1 = φ2 .
The difference in waves is
∆Φ = Φ2 − Φ1 = k (x2 − x1 ) + (x2 − x1 ) − π = k∆x − π.
Now, in order to have constructive interference, the net phase shift has to be a
whole multiple of 2π, and so ∆Φ = k∆x − π = 2mπ, where m = 0, 1, 2, · · · . So,
if k = 2π/λ, and ∆x = 2d, we find
λ=
2d
.
m + 12
Now, here λ is the wavelength in the in the soap, which differs from the wavelength
in the air by a factor of n, λsoap = λanir . So, we can solve for the wavelengths in
air that we would see
2nd
2.66d
λair =
,
1 =
m+ 2
m + 12
since n = 1.33.
(b) Only m = 1 and 2 gives a reflected wavelength in the visible spectrum. For
m = 1, λ = 692 nm, which is reddish, while for m = 2, λ = 415 nm, which is
violet. Together, they look purplish.
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7. Chapter 21 - Problem 77.
Two loudspeakers emit 400 Hz notes. One speaker sits on the ground. The other
speaker is in the back of a pickup truck. You hear eight beats per second as the truck
drives away from you. What is the truck’s speed?
————————————————————————————————————
Solution
Since the truck is driving away from you, it’s frequency is Doppler-shifted. The expression for the shifted frequency that we get is
fobs =
fsource
,
1 + vs /v
where fsource is the source frequency, vs is the velocity of the source, and v = 343 m/s
is the speed of sound.
The beat frequency is the difference between the two frequencies, fbeat = fsource − fobs .
So, plugging in for the source frequency to the Doppler formula gives
fobs =
fobs − fbeat
,
1 + vs /v
Solving for the velocity of the source gives
vs =
fbeat
8
v=
× 343 ≈ 7 /m/s.
fsource
400
So, the truck is driving away from us at about 7 m/s, or about 16 mph.
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8. Chapter 21 - Problem 78.
(a) The frequency of a standing wave on a string is f when the string’s tension is T .
If the tension is changed by the small amount ∆T , without changing the length,
show that the frequency changes by an amount ∆f such that
∆f
1 ∆T
=
.
f
2 T
(b) Two identical strings vibrate at 500 Hz when stretched with the same tension.
What percentage increase in the tension of one of the strings will cause five beats
per second when both strings vibrate simultaneously?
————————————————————————————————————
Solution
q
q
T
1
(a) The speed of a wave on a string is v = µ . Since v = λf , then f = λ Tµ . Now,
taking the derivative,
df
dT
gives
d
df
=
dT
dT
1
λ
s !
T
1 1
√
=
µ
2 2 µT
Now, divide this answer by f to find
r
1 df
µ1 1
11
√
=λ
=
.
f dT
T 2 µT
2T
Separating the derivative gives
df
1 dT
=
.
f
2 T
Letting the infinitesimal differentials become finite deltas, d → ∆, gives us our
answer.
Incidentally, note that this derivation is general - whenever you have some function
y ∼ xα , where α is some number, you can always write dy
= α dx
, which you can
y
x
easily check by integration.
(b) If the difference in frequency, ∆f = 5 beats, and f = 500 Hz, then ∆T
= 2 ∆f
=
T
f
2×5
= 0.02, so a 2% increase in the tension will increase the frequency by 5 beats
500
per second.
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9. Chapter 21 - Problem 80.
As the captain of the scientific team sent to Planet Physics, one of your tasks is to
measure g. You have a long, thin wire labeled 1.00 g/m and a 1.25 kg weight. You have
your accurate space cadet chronometer but, unfortunately, you seem to have forgotten
a meter stick. Undeterred, you first ind the midpoint of the wire by folding it in half.
You then attach on end of the wire to the wall of your laboratory, stretch it horizontally
to pass over a pulley at the midpoint of the wire, then tie the 1.25 kg weight to the
end hanging over the pulley. By vibrating the wire and measuring time with your
chronometer, you find that the wire’s second harmonic frequency is 100 Hz. Next,
with the 1.25 kg weight still tied to one end of the wire, you attach the other end to
the ceiling to make a pendulum. You ind that the pendulum requires 314 seconds to
complete 100 oscillations. Pulling out your trusty calculator, you get to work. What
value of g will you report back to headquarters?
————————————————————————————————————
Solution
We don’t know the length of the string. However, the length of the vibrating part of
the wire is half the total distance, so ` = L/2, where L is the full length of the string.
Now, since the wire is vibrating in the second harmonic, then λ = L2 is the wavelength
p
on the wire. Since λf = v, and v = T /µ, where T = mg is the tension, we can write
r
r
L
1 mg
2 mg
λ= =
⇒L=
.
2
f
µ
f
µ
Now, when we connect the wire
q to the ceiling and let it oscillate as a pendulum, it’s
T 2
oscillation period is T = 2π Lg , which gives L = 2π
g. Setting this equal to our
value for L above gives
2
4
r
T
2 mg
4 2π
m
=
g⇒g= 2
.
f
µ
2π
f
T
µ
Plugging in all the numbers gives g = 8.00 m/s2 , which is a little bit smaller (about
20%) smaller than Earth’s gravity.
9
10. Chapter 21 - Problem 83.
A water wave is called a deep-water wave if the water’s depth is more than one-quarter
of the wavelength. Unlike the waves we’ve considered in this chapter, the speed of a
deep-water wave depends on its wavelength:
r
gλ
.
v=
2π
Longer wavelengths travel faster. Let’s apply this to standing waves. Consider a diving
pool that is 5.0 m deep and 10.0 m wide. Standing water waves can set up across the
width of the pool. Because water sloshes up and down at the sides of the pool, the
boundary conditions require antinodes at x = 0 and x = L. Thus a standing sound
water wave resembles a standing sound wave in an open-open tube.
(a) What are the wavelengths of the first three standing-wave modes for water in the
pool? Do they satisfy the condition for being deep-water waves? Draw a graph
of each.
(b) What are the wave speeds for each of these waves?
(c) Derive a general expression for the frequencies fm of the possible standing waves.
you expression should be in terms of m, g, and L.
(d) What are the oscillation periods of the first three standing-wave modes?
————————————————————————————————————
Solution
The standing wave wavelengths are given by λm =
2L
, where m = 1, 2, 3, · · · . If L = 10.0 m, then this
m
gives λ1 = 20 m, λ2 = 10 m, and λ3 = 6.67 m.
The graphs for each standing wave is seen to the
(a)
right. In order to be a deep-water wave, the depth
has to be more than 1/4 of the wavelength. The
first standing wave just barely fits this condition,
while the other two certainly do.
(b) The wave speeds are calculated from vm =
v2 = 3.95 m/s, and v3 = 3.22 m/s.
10
q
gλm
,
2π
and give v1 = 5.59 m/s,
(c) Because λm fm = vm for each wave, then fm =
since λm =
2L
,
m
1
v
λm m
=
1
λm
q
gλm
2π
=
q
g
.
2πλm
then we can write
r
fn =
ng
,
4πL
where we have redefined m → n to avoid confusion with mass.
(d) Recall that the period of an oscillation is defined as T = 1/f , and so
s
4πL
Tn =
.
ng
Plugging in the values gives T1 = 3.58 s, T2 = 2.53 s, and T3 = 2.07 s.
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Now,