Physics 18 Spring 2011 Homework 13 - Solutions Wednesday April ...

Physics 18 Spring 2011
Homework 13 - Solutions
Wednesday April 20, 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 27th. 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. A string fixed at both ends resonates at a fundamental frequency of 180 Hz. Which of
the following will reduce the fundamental frequency to 90 Hz?
(a) Double the tension and double the length.
(b) Halve the tension and keep the length and the mass per unit length fixed.
(c) Keep the tension and the mass per unit length fixed and double the length.
(d) Keep the tension and the mass per unit length fixed and halve the length.
————————————————————————————————————
Solution
The frequency is given in terms of the wavelength and velocity as f = v/λ. Now the
speed of the wave depends on the tension and mass per unit length as v = T/µ, and
so
f = 1
λ
T
µ .
Now, the fundamental frequency of the wave has a wavelength of 2L, where L is the
length of the string. So,
f1 = 1
2L
T
µ .
If we want to make the frequency half of what it was, f1 → f1 , then we should follow
2
method (c), where we keep the tension and mass per unit length fixed, and double the
length, which will give
f = 1
4L
T
µ = 1
2
1
2L
T
µ
= f1
2 .
1

2. Estimate the resonant frequencies that are in the audible range of human hearing of
the human ear canal. Treat the canal as an air column open at one end, stopped at
the other end, and with a length of 1.00 in. How many resonant frequencies lie in
this range? Human hearing has been found experimentally to be the most sensitive
at frequencies of about 3, 9, and 15 kHz. How do these frequencies compare to your
calculations?
————————————————————————————————————
Solution
The ear is an open-closed tube, and so has allowed standing-wave frequencies of fn =
v
4Ln = f1n, where n = 1, 3, 5, · · · , and f1 = v 343 = = 3376 Hz, or 3.376 kHz, is
4L 4×0.0254
the fundamental frequency (we have recalled that one inch is 2.54 centimeters. The
next few frequencies are
f3 = 3f1 = 3(3.376) = 10.13 kHz
f5 = 5f1 = 5(3.376) = 16.9 kHz
f7 = 7f1 = 7(3.376) = 23.6 kHz.
Now, the high-end of audible frequencies for the human ear is about 20 kHz, so the
n = 7 mode isn’t within range. So, there are only three resonant modes, n = 1, 3, 5,
and the theoretical values are fairly close to the experimental values.
2

3. It is thought that the brain determines the direction of the source of a sound by sensing
the phase difference between the sound waves striking the eardrums. A distance source
emits sound of frequency 680 Hz. When you are directly facing a sound source there
is no phase difference. Estimate the phase difference between the sounds received by
your ears when you are facing 90 ◦ away from the direction of the source.
————————————————————————————————————
Solution
The phase difference is just due to the extra distance that one wave travels over another,
δ = 2π∆x.
We can rewrite this in terms of the speed and frequency of the wave, v = λf,
λ
giving
δ = 2π
v ∆xf.
Suppose that there is a distance of about 20 centimeters between your ears. Then,
since the speed of sound is 343 m/s, we have
δ = 2πf ∆x
v
= 2π (680) .2
343
3
= 2.49 rad.

4. A standing wave on a rope is represented by the wave function
1
y(x, t) = (0.020) sin
2 πx
cos(40πt),
where x and y are in meters, and t is in seconds.
(a) Write wave functions for two traveling waves that, when superimposed, produce
this standing-wave pattern.
(b) What is the distance between the nodes of the standing wave?
(c) What is the maximum speed of the rope at x = 1.0 m?
(d) What is the maximum acceleration of the rope at x = 1.0 m?
————————————————————————————————————
Solution
(a) Recall that the sum of two sine waves is
1
1
sin A + sin B = 2 cos (A − B) sin (A + B) .
2 2
This is precisely the correct form of the given standing wave, so we can write
π
π
y (x, t) = 0.010 sin x − 40πt + 0.010 sin x + 40πt ,
2 2
which gives the superposition of a wave traveling to the right and another traveling
to the left.
(b) The distance between the nodes is just half the wavelength. Now, the wave vector
k = 2π/λ = π/2, so λ = 4 meters, and thus the distance between the nodes is
d = 2 meters.
(c) The speed of the rope (not the wave) is just the derivative of the displacement,
v = ˙y. Taking the derivative gives
v = d
dt
(0.020) sin 1
2 πx cos(40πt)
= (0.020) sin 1
2 πx d
dt [cos(40πt)]
= −(0.800)π sin 1
2 πx sin(40πt).
Now, the maximum speed occurs when the cosine is −1 (to cancel the minus sign
in the velocity). So, the maximum velocity is v = 0.8π sin π
2 x . When x = 1,
then
π
v (x = 1) = 0.8π sin = 0.8π m/s = 2.5 m/s.
2
(d) The acceleration is just the second derivative of the position, a = ¨y, or
a = d2
dt 2
(0.020) sin 1
2 πx cos(40πt)
= (0.020) sin 1
2πx d2 dt2 [cos(40πt)]
= −32π2 sin 1
2πx sin(40πt).
Again, the maximum acceleration occurs when the cosine is −1, and when x = 1,
we find amax = 32π2 m/s2 .
4

5. Ultrasound has many medical applications, one of which is to monitor fetal heartbeats
by reflecting ultrasound off a fetus in the womb.
(a) Consider an object moving at speed v0 toward an at-rest source that is emitting
sound waves of frequency f0. Show that the reflected wave (i.e., the echo) that
returns to the source has a Doppler-shifted frequency
v + v0
fecho = f0,
v − v0
where v is the speed of sound in the medium.
(b) Suppose that the object’s speed is much less than the wave speed: v0 ≪ v. then
fecho ≈ f0, and a microphone that is sensitive to these frequencies will detect
a beat frequency if it listens to f0 and fecho simultaneously. Use the binomial
expansion and other appropriate approximations to show that the beat frequency
is fbeat ≈ 2v0
v f0.
(c) The reflection of 2.40 MHz ultrasound waves from the surface of a fetus’s beating
heart is combined with the 2.40 MHz wave to produce a beat frequency that
reaches a maximum of 65 Hz. What is the maximum speed of the surface of the
heart? The speed of ultrasound waves within the body is 1540 m/s.
(d) Suppose the surface of the heart moves in simple harmonic motion at 90 beats/min.
What is the amplitude in mm of the heartbeat?
————————————————————————————————————
Solution
(a) The source is at rest and is sending out the waves, which leads to a (higher)
Doppler-shifted frequency received by the object, fR = f0 (1 + v0/v). This frequency
is reflected back to the source, but the object now looks like a moving
source, and the frequency is Doppler-shifted for a second time, fecho = fR
1−v0/v .
Plugging in for fR from before gives
1 + v0/v v + v0
fecho =
f0 = f0,
1 − v0/v v − v0
as claimed.
(b) The beat frequency is just the difference between the echo frequency and the
originalo frequency, fecho − f0. Now, recalling that v0 ≪ v, then, recalling that
(1 + ɛ) α ≈ 1 + αɛ, for small ɛ,
1+v0/v
fecho =
1−v0/v
1 − v0
= 1 + v0
v
≈ 1 + v0
v0
1 + v
2v0
≈ 1 + f0,
5
v
f0
−1 f0 v
f0 v

where we have dropped terms of order (v0/v) 2 and higher. Thus,
fbeat ≈ fecho − f0 = 1 + 2v0
f0 − f0 =
v
2v0
v f0.
(c) Here we just plug in the numbers to find v0 using our results from part (b),
or about 2.1 cm/s.
v0 = fbeat
v =
2f0
65
× 1540 = 0.0209 m/s,
2 × 2.40 × 106 (d) The amplitude, A, is given in terms of the maximum velocity as vmax = v0 = ωA,
where ω = 2πf is the angular frequency. So,
which is about 2.2 mm.
A = v0
2πf
= 0.0209
2π (90/60)
6
= 0.0022 m,