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course Phy 202
2/23 10:50 110221 Physics
Intro Set 6: http://vhmthphy.vhcc.edu/ph2introsets/default.htm
`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.
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What is wavelength of the wave?
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The string is 3/2 of a cycle, so 3/2 `lambda=6 m so `lambda=4 m
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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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C=30 m/s then f=7.5 cycles/s
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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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f=c/`lambda f=340 m/s/3.2 m=106.25 cycles/sec
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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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Six longest are 7.2 m, 3.6 m, 2.4 m, 1.8 m, 1.44 m, and 1.2 m
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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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c=f*`lambda c=30 cycles/s*7.2 m=216 m/s
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What then are the frequencies of the other five wavelengths?
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Respectively: f=c/`lambda
60 cycles/s, 90cycles/sec, 120 cycles/sec, 150 cycles/sec, 180 cycles/sec
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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.
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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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3/4 `lambda
5/4 `lambda
7/4 lambda
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How long is the pipe? Recall that sound in air travels at about 340 m/s.
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1/4`lambda=340 m/s/ 20 cycles/sec= 17 m then the pipe is 68 m (4*17 meters)
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@& The node-antinode configurations are
N A
N A N A
N A N A N A.
As shown in an earlier problem we conclude that the wavelengths are respectively 4, 4/3 and 4/5 the length of the pipe.
The frequencies are therefore
c / (4 L), where c is the speed of propagation and L the length of the pipe,
c / (4/3 L) = 3 * c / (4 L)
c / (4/5 L) = 5 * c / (4 L).
The first frequency is 20 Hz, and c = 340 m/s, so
340 m/s / (4 L) = 20 Hz, and it follows that
4 L = 340 m/s / (20 cycles / sec) = 17 meters
so that
L = 17/4 meters.
The other frequencies are respectively 3 and 5 times as great as the first, so the frequencies are 3 * 20 Hz = 60 Hz and 5 * 20 Hz = 100 Hz.
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@& Very good. See my notes.*@