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course Phy 242
March 8 around 2:30pm.
110221`q001. A standing wave in a string has three antinodes between its ends, and the ends are nodes. The ends are separated by 6 meters.
What is wavelength of the wave?
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(3/2)(lam) = 6m
Lam = (2/3)*(6) = 4 meters
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If the propagation velocity in the string is 30 meters / second, what is the frequency of the wave (i.e., how many complete cycles occur per second)?
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(30m)/(4m/cycle) = 7.5 cycles in 30 m
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`q002. A standing sound wave in a pipe 4 meters long has an antinode at one end and a node at the other. Nodes and antinodes are equally spaced, and two antinodes occur within the pipe. Sound waves propagate at about 340 meters / second. What is the frequency of this wave?
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(340m/s)/(16m/cycle) = 21.25 Hz or (cycle/sec)
@& This would be true for the fundamental frequency, with a single antinode at the end.
If there are two nodes within the pipe, then the wavelength is shorter than the fundamental.
The minimal configuration of nodes and antinodes that meets the given conditions is
N A.
In this configuration the length of the pipe would be 1/4 wavelength, so the 1/4 wavelength = 4 meters and the wavelength would be 4 * 4 meters = 16 meters.
If an antinode occurs within the pipe, then the configuration is
N A N A
consisting of three quarter-wavelengths. So
3/4 * wavelength = pipe length
and
wavelength = 4/3 * pipe length
If two antinodes occur within the pipe, then the configuration must be
N A N A N A,
consisting of five quarter-wavelengths. So
5/4 * wavelength = pipe length
and
wavelength = 4/5 * pipe length.
Note the progression of wavelengths: 4/1 * pipe length, 4/3 * pipe length, 4/5 * pipe length.
You should see that the denominators correspond to the number of node-antinode distances, and that this number increases by 2 from one configuration to another, due to the fact that nodes and antinodes must alternate.
This will always be the pattern for a situation in which a node occurs at one end, with an antinode at the other.
For the situation of the question, the pipe length is 4 meters, so the wavelength is 4/5 * 4 meters = 3.2 meters.
In 340 meters there would therefore be
340 m / (3.2 m / wavelength) = 106.25 wavelengths,
each representing a cycle, so the frequency would be
f = 106.25 cycles / second.
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`q003. A string has nodes at both of its ends, which are 3.6 meters apart. It is theoretically possible to achieve a standing wave of any wavelength, provided that nodes and antinodes alternate at equal spacing.
What therefore are the six longest wavelengths that can occur in this string?
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7.2, 3.6, 2.4, 1.8, 1.44, 1.2
(ex: 2(1/4)*lam = 3.6)
(lam = 7.2)
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If the first of these has a frequency of 30 cycles / second, what is the speed of propagation in the string?
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F = (30 cycle/sec)
Lam = 7.2
C= (30 cycle/sec) * (7.2)
= 216 m/s
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What then are the frequencies of the other five wavelengths?
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Would I just do like I did above for each of the wavelengths?
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`q004. A sound wave in a pipe has a node at one end and an antinode at the other. The frequency of the longest possible wave in this pipe is 20 Hz.
What are the three next-lowest frequencies possible in this pipe? Once more, nodes and antinodes must alternate and must be equally spaced.
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N A
N A N A
N A N A N A
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How long is the pipe? Recall that sound in air travels at about 340 m/s.
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340m/s = 20(lam)
Lam = 17m
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@& You've got the wavelengths and the speed of propagation, so you can reason out the frequencies as you have done before.
For reference:
The six configurations having nodes at both ends, corresponding to the six shortest wavelengths are
N A N
N A N A N
N A N A N A N
N A N A N A N A N
N A N A N A N A N A N
N A N A N A N A N A N A N
having 2, 4, 6, 8, 10 and 12 quarter-wavelengths, respectively.
Thus we have
2 * 1/4 wavelength = string length
4 * 1/4 wavelength = string length
6 * 1/4 wavelength = string length
...
...
12 * 1/4 wavelength = string length
so that wavelengths can be 4/2, 4/4, 4/6, 4/8, 4/10 and 4/12 of the string length.
These fractions could of course be reduced, but they were left unreduced to show how the denominators increase by 2 with each new configuration, corresponding to the addition at each step of a node-antinode pair with its two quarter-cycles.
Compare with the denominators that occur for the wavelengths of the pipe which is closed on one end and open on the other.
String length is 3.6 meters, so the wavelengths work out to be 7.2, 3.6, 2.4, 1.8, 1.44 and 1.2 meters.
If the first has a frequency of 30 cycles / sec, then the propagation velocity is
30 cycles / sec * 7.2 m / cycles = 216 m/s.
The corresponding frequencies of the other harmonics are
216 m/s / 3.6 m = 60 cycles / sec
216 m/s / 2.4 m = 90 cycles / sec
216 m/s / 1.8 m = 120 cycles / sec
216 m/s / 1.44 m = 150 cycles / sec
216 m/s / 1.2 m = 180 cycles / sec.
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