Physics 1240 Website: http://www.colorado.edu/physics/phys1240
Homework 3 Solutions
Physics 1240 Spring 2005
In all homework problems where there are calculations, you should show the steps
you took in arriving at your result: the correct answer alone is not enough! Also be
sure to include units with any results, e.g. “f1 = 440 Hz” (not just “f1 = 440”).
1. What are the frequencies of the first and third harmonics of an open cylindrical
pipe 2 m long? What are the frequencies of the fundamental and third harmonic if
one end of the pipe is blocked? Sketch the standing wave patterns for the third
harmonic in each case, showing both the pressure and displacement, as in Figure
12.2. (4 pts)
For a 2m open pipe:
f 1 = v/ λ = v/2L
f3 = 3*f1
= 344/(2*2)
= 86 Hz
= 3*86
= 258 Hz
For a 2m closed pipe:
f 1 = v/ λ = v/4L = 344/(4*2)
= 43 Hz
f 3 = 3*f1
= 3*43
= 129 Hz
1
Physics 1240 Website: http://www.colorado.edu/physics/phys1240
2. A clarinet and a flute both have nearly cylindrical bores with similar length and
diameter, yet they are quite different in both tone quality and range of pitch. Discuss
these differences more specifically, explaining the underlying reasons. (2 pts)
The mouthpiece of the flute is blown across, but not covered by the mouth. This
effectively makes the flute act as a pipe open at both ends. A resonating pipe
open at both ends contains all integer multiple harmonics of the fundamental
frequency. The clarinet, on the other hand, is nearly cylindrical like the flute,
but the mouth/mouthpiece act as a stop on one end, effectively creating a pipe
that is closed at one end. Pipes closed at one end only produce odd multiple
harmonics of the fundamental frequency.
3. The clarinet’s lowest note is D3, 147 Hz. What is its effective length, i.e. the length
of an idealized, closed pipe that has this frequency for its first mode? (2 pts)
In addition to the tonal differences of the harmonic content, the fundamental
frequency of the clarinet will be half of the flute. A “stopped” pipe, or closed at
one end, drops the fundamental frequency in half, like in the first problem.
L = 1/4λ => λ = 4L
f = v/ λ = v/4L => L = v/4f
L = v/4f
=344/(4*147)
= 0.585 m
2
Physics 1240 Website: http://www.colorado.edu/physics/phys1240
4. To obtain a fundamental vibration frequency of 860 Hz, what length open tube
would be required? What length of tubing closed at one end? (2 pts)
Open
L = 1/2λ => λ = 2L
f = v/ λ = v/2L => L = v/2f
L = v/2f
=344/(2*860)
= 0.2 m
Closed
L = 1/4λ => λ = 4L
f = v/ λ = v/4L => L = v/4f
L = v/4f
=344/(4*860)
= 0.1 m
5. Suppose your flute has a leaky keypad about halfway down its length. How will
this affect the instrument’s performance? (2 pts)
A flute with a leaky keypad will allow air to escape, creating a shorter effective
length of pipe. The pitch will therefore go up (in principle by about an octave).
6. How does a trumpet player obtain notes with different pitch? (2 pts)
A trumpet player obtains notes with different pitches both by using valves,
which change the effective length of the tubing in the horn, and by tightening
the lips and blowing harder to excite higher modes.
7. Why do we always model reed and brass instruments as “pipes with one closed
end” when in fact you have to let air in at the mouthpiece in order to get any sound?
(2 pts)
Brass and reed instruments act as “pipes with one end closed” because the
mouth/mouthpiece is mostly closed except to let in small puffs of air. The
mouthpiece end therefore acts as a barrier, and the only open side of the
instrument is the bell.
3
Physics 1240 Website: http://www.colorado.edu/physics/phys1240
8. Assume that the length of a trombone is 275 cm in first (“open”) position
(indicated below). How far should the slide be moved to lower the pitch one
semitone? (Remember the tube length is increased by twice this amount.) In the equal
tempered tuning system, a semitone difference in pitch corresponds to 6% difference
in frequency. (4 pts)
L = 1/4λ => λ = 4L
f = v/ λ = v/4L
= 344/(4*2.75)
= 31.3 Hz
A pitch 6% lower in frequency would be
94% of 31.3 Hz = 29.4 Hz
Now working backwards to find the tube length for the new frequency:
f = v/ λ = v/4L => L = v/4f
L = v/4f
= 344/(4*29.4)
= 2.93 m, or 293 cm
The difference is 293 cm (new length) – 275 cm (original length) = 18 cm
The tubing length needs to change 18 cm total to lower the pitch by a semitone,
but since the length of the slide of a trombone is doubled (down and back), the
actual slide needs to move out only half of the 18 cm, or 9 cm.
4