Physics 18 Spring 2011 Homework 12 - Solutions Wednesday April ...

Physics 18 Spring 2011
Homework 12 - Solutions
Wednesday April 13, 2011
Make sure your name is on your homework, and please box your final answer. Because
we will be giving partial credit, be sure to attempt all the problems, even if you don’t finish
them. The homework is due at the beginning of class on Wednesday, April 20th. Because
the solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1. The wave function for a harmonic wave on a string is y(x, t) = (1.00 mm) sin(62.8 m −1 x+
314 s −1 t).
(a) In what direction does this wave travel, and what is the wave’s speed?
(b) Find the wavelength, frequency, and period of this wave.
(c) What is the maximum speed of any point on the string?
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Solution
Recall that our usual expression for a wave is
y(x, t) = A sin (kx ± ωt + δ) ,
where A is the amplitude, k = 2π/λ is the wave number, ω = 2πf is the angular
frequency, and δ is a phase shift. We’ll use this form to answer our questions.
(a) Because the coefficient of the time is positive, the wave travels to the left. It’s
speed is just ω/k = 314/62.8 = 5 m/s.
(b) The wavelength is λ = 2π/k = 2π/62.8 = 0.10 meters. It’s frequency is f =
ω/2π = 314/2π = 50 Hz, which gives a period of T = 1/f = 1/50 = 0.02 seconds.
(c) The maximum speed of any point on the string is just the amplitude of the time
derivative of the wave function (the velocity), which is Aω. So, vmax = Aω =
0.001 × 314 = 0.314 m/s.
1

2. You have just been pulled over for running a red light, and the police officer has
informed you that the fine will be $250. In desperation, you suddenly recall an idea
that your physics professor recently discussed in class. In your calmest voice, you tell
the officer that the laws of physics prevented you from knowing that the light was red.
In fact, as you drove toward it, the light was Doppler shifted to where it appeared
green to you. “OK,” says the officer, “Then I’ll ticket you for speeding. The fine is $1
for every 1 km/hr over the posted speed limit of 50 km/hr.” How big is your fine? Use
650 nm as the wavelength of red light and 540 nm as the wavelength of green light.
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Solution
Since we are approaching the the light, the wavelengths are shortened due to the
Doppler shift. The decrease in the wavelength is given by
c − v
λ =
c + v λ0
1 − v/c
=
1 + v/c λ0.
Once again, we can solve for the velocity as
v
c = λ20 − λ2 λ2 .
0 + λ2 Taking λ0 = 650 nm and λ = 540 nm, we find
v
c = λ20 − λ2 λ2 0 + λ2 = 6502 − 5402 6502 = 0.183.
+ 5402 Thus, the velocity is v = 0.183c ≈ 55.5 × 10 7 m/s, which is about 2 × 10 8 km/hr. This
is well above the posted speed limit of 50 km/hr, and so the fine would basically be
$200 million!!! You’re far better off just paying the $250 fine for running the red light!
2

3. In the oceans, whales communicate by sound transmission through the water. A whale
emits a sound of 50.0 Hz to tell a wayward calf to catch up to the pod. The speed of
sound in water is about 1500 m/s.
(a) How long does it take the sound to reach the calf if he is 1.20 km away?
(b) What is the wavelength of this sound in the water?
(c) If the whales are close to the surface, some of the sound energy might refract out
into the air. What would be the frequency and wavelength of the sound in the
air?
————————————————————————————————————
Solution
(a) This is just a simple velocity times time problem, so t = d/v = 1200/1500 = .8
seconds.
(b) Since the velocity is the product of the wavelength and the frequency, then λ =
v/f = 1500/50 = 30 meters.
(c) The frequency doesn’t change as we go from one medium (water) to another
(air), and so the frequency of the wave is still the same 50.0 Hz. Now, the
velocity of the sound wave is much slower, vair = 343 m/s. So, the wavelength is
λ = v/f = 343/50 = 6.86 meters.
3

4. In modern medicine, the Doppler effect is routinely used to measure the rate and
direction of blood flow in arteries and veins. High-frequency “ultrasound” (sound
at frequencies above the human hearing range) is typically employed. Suppose you
are in charge of measuring the blood flow in a vein (located in the lower leg of an
older patient) that returns blood upward to the heart. Her varicose veins indicate that
perhaps the one-way valves in the vein are not working properly and that the the blood
is “pooling” in the veins and perhaps even that the blood flow is backward toward her
feet. Employing sound that has a frequency of 50.0 kHz, you point the sound source
from above her thigh region down toward her feet and measure the sound reflected
from that vein area to be lower than 50.0 kHz.
(a) Was your diagnosis of the valve condition correct? If so, explain.
(b) Estimate the instrument’s frequency difference capability to enable you to measure
speeds down to 1.00 mm/s. Take the speed of sound in flesh to be the same as
that in water, 1500 m/s.
————————————————————————————————————
Solution
(a) Because the reflected frequency was lower than the emitted frequency, we know
that the source was moving away from you. This means that the blood wasn’t
coming back up the veins to the heart, and so your diagnosis is correct.
(b) When we first emit the pulse, the blood is moving away from us. This results in
a shift received by the blood,
fR =
v
fS,
v − uS
where uS is the velocity of the blood. This, then, results in a shifted source
frequency, fR, that is reflected back as us, leading to a second Doppler shift. We
get a new frequency, f ′ R . Since the source is moving away from us, the reflected
frequency is given by the Doppler shift as
f ′ R = vfR
v − uS
=
v2 2 fS,
(v − uS)
So, the difference in frequency is ∆f = f ′ R − fS, or
∆f = 1 − v2
(v−uS) 2 fS
−2vus+u2 s
(v−us) 2 fS
=
−2vus
≈
(v) 2 fS
= −2 us
v fS,
where we have recalled that v ≫ us in the last two lines. So, plugging in the
values gives
∆f = −2 us
v fS = −2 0.001 3
50 × 10
1500
≈ −0.067 Hz.
4

5. A bat flying toward a stationary obstacle at 12.0 m/s emits brief, high-frequency sound
pulses at a repetition frequency of 80.0 Hz. What is the interval between the arrival
times of the reflected pulses heard by the bat?
————————————————————————————————————
Solution
Because the bat is getting closer the frequency with which it receives the reflected
pulses is increasing, meaning that the period between the reception of those pulses is
going down. The time between pulses is just the inverse of the received frequency,
∆t = 1/fR. Now, using the Doppler shift formula we can relate the received frequency
to the source frequency. Since the bat is approaching the “source” (the reflecting wall)
we have
fR =
v + uR
fS,
v − uS
where v is the speed of the wave, uR = uS = 12.0 m/s is the speed of the bat, and fS
is the frequency of the sound emitted by the bat. So, since ∆t = 1/fR, we have
∆t = 1
=
fR
v − uS
1 343 − 12 1
=
= 0.0112 seconds.
v + uR fS 343 + 12 80
5