Solutions to Test 1 Review Problems
1. Here are two sound waves. Identify which is periodic. For the periodic wave, sketch a
fundamental cycle, measure the fundamental period, and calculate the frequency in Hz and
BPM. Is the frequency within the range of human hearing?
Answer. The periodic wave is on the left. Its fundamental cycle looks like this:
and its
fundamental period is 0.01 seconds. Therefore, its frequency is 1/0.01 = 100 Hz = 6000 BPM.
2. Functions.
(a) For each function, determine whether or not it is periodic. If it is periodic, list the
smallest four positive numbers P for which f (t) = f (t + P ), identify the fundamental
period, and calculate the frequency, using appropriate units in your answers.
i. f (0) = 10, f (1) = 20, f (2) = 10, f (3) = 20, …
ii. f (t) = 3 sin(1000πt)
iii. f (t) = t2
Answer.
i. periodic with P = 2, 4, 6, 8, . . ., fundamental period = 2 seconds, frequency = 0.5 Hz.
ii. periodic with P = 0.002, 0.004, 0.006, . . ., fundamental period = 0.002 seconds,
frequency = 500 Hz.
iii. not periodic.
(b) Find the amplitude, fundamental period, and frequency of each sinusoid, and sketch a
fundamental cycle: (1) y = 3 sin(1000πt)
(2) y = 2 sin(0.31415t).
Answer. (1) Amplitude = 3, fundamental period = 2π/(1000π) = 0.002, frequency =
500 Hz. (2) Amplitude = 2, fundamental period = 2π/0.31415 ≈ 20 seconds, frequency =
0.05 Hz.
3. Write the formula for a sine wave y = A sin(Bt), where t is measured in seconds, with (1)
amplitude = 5 and frequency =3 BPM
(2) amplitude = 10 and fundamental period = 0.005 seconds.
Answer. (1) since 3 BPM = 180 Hz, the formula is 5 sin(360πt) (2) since frequency =
1/fundamental period = 200 Hz, the formula is 10 sin(400πt)
4. Beats.
(a) Suppose a 1000 Hz tone is modulated by the following envelope. Sketch the resulting
sound wave (what you would see in Audacity) and describe in words what you would
hear. Envelope =
Answer. Sound wave =
You hear a sound that fades in, then
gets quiet, then gets louder, and inally fades out. The width of the sound wave
corresponds to intensity.
(b) Suppose a 1000 Hz tone and a 1008 Hz tone are played simultaneously. Describe what
you would hear, and make a sketch of the irst second of the resulting sound ile.
Answer. You hear a 1004 Hz tone pulsing with beats 8 times a second.
5. Describe how the phenomenon of beats can be used to produce an 1000 Hz tone that pulses at
a rate of 120 BPM.
Answer. Since 120 BPM = 120/60 = 2 Hz, we need to play a 999 Hz tone and a 1001 Hz tone
simultaneously.
6. Suppose a willow lute has a fundamental frequency of 400 Hz when played with the end open.
Calculate the frequencies of the second and third harmonics and sketch the standing waves
corresponding to these harmonics. Describe how a 600 Hz tone could be produced on the
same lute.
Answer. Harmonics with end open: 400, 800, 1620, etc. Harmonics with end closed: 200, 600,
1000, 1400, etc. So the second and third harmonics are 800 and 1620 Hz, and you can get a
600 Hz tone by closing one end. Here are the standing waves of the irst three harmonics:
7. Suppose that when a slide whistle has an inside length of 6 inches, its irst harmonic has
frequency 800 Hz. If the length is decreased to 5 inches, what is the frequency of the irst
harmonic that results? What length would produce a sound that is an octave above the original
frequency?
Answer. Use the formula frequency = an/(2L), where a is a physical constant, n is the number
of the harmonic, and L is the length of the tube. So when L = 6 and n = 1, frequency = 800 =
a/12, so a = 9600. Since a is a number that doesn’t change, we can plug it into the equation and
use L = 5. Then frequency = 9600/10 = 960 Hz. If L = 3, then frequency = 9600/6 = 1600 Hz,
which is an octave about 800 Hz.
8. Which picture represents a harmonic spectrum? Label the units on the axes. If one picture is
the spectrum of a drum and the other is the spectrum of a violin, which is which?
Answer. A harmonic spectrum (like that of a violin or lute) has equally-spaced peaks, so the
drum is on the left and the violin is on the right. The horizontal axis on the graphs is frequency
(in Hz) and the vertical axis is intensity (in dB).
9. This picture is a spectrogram. Label the units on the axes, and describe the sound that will be
heard.
Answer. The horizontal axis is time in seconds and the vertical axis is frequency in Hz. You will
hear a tone that gets lower for 30 seconds, then climbs to a higher frequency than where it
started.
10. Find the following frequencies: (1) an octave above 600 Hz (2) an octave below 600 Hz (3) a
perfect ifth above 600 Hz (4) a perfect ifth below 600 Hz. Explain why two lutes playing
400 Hz and 800 Hz sound better than lutes playing 400 Hz and 820 Hz.
Answer. (1) 1200 Hz; (2) 300 Hz; (3) 900 Hz; (4) 400 Hz. If lute A plays 400 Hz and lute B
plays 800 Hz, then every frequency in lute B’s spectrum is also in lute A’s spectrum, so the
lutes sound good. If lute A plays 400 Hz and lute B plays 820 Hz, there will be a peak in lute
A’s spectrum at 800 Hz, which causes beats when played with an 820 Hz sound. Beats sound
“rough” and not harmonious.