Phys 197 Homework Solution 15A
Q12.
The four strings on a violin have different thicknesses, but are all under approximately
the same tension. Do waves travel faster on the thick strings or the thin strings? Why? How does
the fundamental vibration frequency compare for the thick versus the thin strings?
————–
p
Since v = F/µ, where F is the tension and µ is the linear mass density (in kg/m), waves
travel faster on the thin strings. The period is proportional to the time for a wave to travel the
length of the string, so the thin strings have a shorter period and hence a higher frequency.
Q21.
A musical interval of an octave corresponds to a factor of 2 in frequency. By what factor
must the tension in a guitar or violin string be increased to raise its pitch one octave? To raise it two
octaves? Explain your reasoning. Is there any danger in attempting these changes in pitch?
————–
p
F0 /µ, v1 = 2v0 =
Doubling
the
frequency
calls
for
doubling
the
wave
speed.
Since
v
0 =
p
p
2 F0 /µ = 4F0 /µ. The tension must be increased to 4F0 , a factor of 4. Going up two octaves
would be a factor of 16 in tension. Sounds like a good chance of broken strings or other parts.
P1.
The speed of sound in air at 20◦ C is 344 m/s (a) What is the wavelength of a sound wave
with a frequency of 784 Hz, corresponding to the note G5 (the G just above the treble staff) on a
piano, and how many milliseconds does each vibration take? (b) What is the wavelength of a sound
wave one octave higher than the note in part (a)?
————–
(a) λ = v/f = (344 m/s)/784 Hz) = 0.439 m.
T = 1/f = 1/(784 Hz) = 0.001276 s = 1.276 ms.
(b) The wavelength is halved: λ = 0.219 m.
P4.
Ultrasound Imaging. Sound having frequencies above the range of human hearing (about
20,000 Hz) is called ultrasound. Waves above this frequency can be used to penetrate the body and
to produce images by reflecting from surfaces. In a typical ultrasound scan, the waves travel through
body tissue with a speed of 1500 m/s. For a good, detailed image, the wavelength should be no more
than 1.0 mm. What frequency sound is required for a good scan?
————–
Since v = λf , f = v/λ = (1500 m/s)/(10−3 m) = 1.5 MHz. That’s the same frequency as
“1500” on your AM radio dial (in sound waves, versus electromagnetic waves).
P7.
Transverse waves on a string have wave speed 8.00 m/s, amplitude 0.0700 m, and wavelength
0.320 m. The waves travel in the −x-direction and at t = 0 the end of the string has its maximum
upward displacement. (a) Find the frequency, period, and wave number of these waves. (b) Write a
wave function describing the wave. (c) Find the transverse displacement of a particle at x = 0.360 m
at time t = 0.150 s. (d) How much time must elapse from the instant in part (c) until the particle at
x = 0.360 m next has maximum upward displacement?
————–
(a) f = v/λ = (8 m/s)/(0.32 m) = 25.0 Hz. T = 1/f = 0.0400 s. k = 2π/λ = 19.64 m−1 .
(b) The standard formula for a wave is y(x, t) = A cos(kx ± ωt). We have A and k, but still
need ω.
ω = 2πf = (6.28 (rad/s)/Hz)(25 Hz) = 157.1 rad/s. Note the units I ascribe to the 2π factor.
y(x, t) = (0.0700 m) cos(19.64 x + 157.1 t). (Suppressing the units). The plus sign is chosen
to correspond to the wave traveling in the negative x direction.
(c) Plug: y(0.36 m, 0.15 s) = (0.7 m) cos(7.0704 + 23.565) = 0.0497 m. Keep extra sigdigs.
(d) Calculate the total phase (argument of the cosine), which is of 7.0704 rad + 23.565 rad =
30.635 rad. Divide this by 2π to see how many cycles have elapsed:
Number of cycles = (30.635 rad)/(2π) = 4.8757. So it needs just 0.1243 of a period to
re-attain max:
0.1243 T = 0.1243 · (0.0400 s) = 0.0497 s.
This differs slightly from the answer in the back of the book due to the extra sigdig. It shows
that a very small error (or change) in k or ω accumulates over a number of cycles.
P8.
A certain transverse wave is described by
y(x, t) = (6.50 mm) cos 2π
³
´
t
x
−
28.0cm 0.0360 s
Determine the wave’s (a) amplitude; (b) wavelength; (c) frequency; (d) speed of propagation; (e)
direction of propagation.
————–
(a) Amplitude is absolute value of coefficient: A = 6.50 mm.
(b) Wavelength: We know that k = 2π/λ in general, and that k = 2π/(28.0 cm) in this instance.
So, λ = 28.0 cm.
(c) Frequency: Similarly, ω = 2π/T in general, and ω = 2π/(0.0360 s) in this instance, so
T = 0.360 s. Frequency is the reciprocal:
f = 1/T = 1/(0.360 s) = 27.8 Hz.
(d) Speed of propagation: v = λ/f = (0.28 m)/(0.0360 s) = 7.78 m/s. Converted to m/s as a
matter of habit, in case it gets plugged into something else.
(e)Direction of propagation: +x, since there is a negative sign between kx and ωt.
P15.
One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that
vibrates the rope transversely at 120 Hz. The other end passes over a pulley and supports a 1.50-kg
mass. The linear mass density of the rope is 0.0550 kg/m. (a) What is the speed of a transverse wave
on the rope? (b) What is the wavelength? (c) How would your answers to parts (a) and (b) change
if the mass were increased to 3.00 kg?
————–
p
p
p
(a) v = F/µ = (1.5 kg)(9.8 m/s2 )/(0.0550 kg/m) = √267.3 m2 /s2 = 16.3 m/s. If you
forgot to multiply by g, the units on the RHS would be m; since this is not the unit for
velocity, you would know to double-check.
(b) v = λf ⇒ λ = v/f = (16.3 m/s)/(120
Hz) = 0.136 m.
√
(c) v would increase by a factor of 2 = 1.414 to 23.1 m/s and wavelength would increase by
the same factor to 0.192 m.
P42.
A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m
long and has a mass of 3.00 g. (a) What is the frequency of its fundamental mode of vibration? (b)
What is the number of the highest harmonic that could be heard by a person who is capable of hearing
frequencies up to 10,000 Hz?
————–
(a) The fundamental is f1 of Eq 15.32: f1 = v/2L. The v in the equation is the speed of the
wave on
p given by
pthe string,
F/µ
=
(800 N)/[(0.003 kg)/(0.4 m)] = 326.6 m/s.
v=
Note that converting grams to kg is essential.
f1 = v/2L = (326.6 m/s)/(2 · 0.4 m) = 408.3 Hz.
(b) 10 000 Hz/408.3 Hz = 24.5, so up to the 24th harmonic at 9800 Hz.
P49.
Guitar String. One of the 63.5-cm-long strings of an ordinary guitar is tuned to
produce the note B3 (frequency 245 Hz) when vibrating in its fundamental mode. (a) Find the speed
of transverse waves on this string. (b) If the tension in this string is increased by 1.0%, what will
be the new fundamental frequency of the string? (c) If the speed of sound in the surrounding air is
344 m/s find the frequency and wavelength of the sound wave produced in the air by the vibration
of the string. How do these compare to the frequency and wavelength of the standing wave on the
string?
————–
(a) f1 = v/2L, so v = 2Lf1 = 2 · (0.635 m)(245 Hz) = 311 m/s. Note the necessity to convert
centimeters to meters.
√
(b) If tension is multiplied by 1.01, the speed v is to be multiplied by 1.01 = 1.005. The
fundamental is directly proportional to the velocity, so
f1′ = 1.005 f1 = (1.005)(245 Hz) = 246.2 Hz.
(c) Key point is that the frequency is the physical shaking, which has to be the same. So:
fin air = 245 Hz.
The fundamental on the string is such that the length of the string is one-half a wavelength:
λon string = 2L = 1.270 m.
In air, use the formula v = λf :
λ = v/f = (344 m/s)/(245 Hz) = 1.404 m.