A pipe is open at both ends. The pipe has resonant frequencies of 655 Hz and 786 Hz (among others). Find the two lowest possible values for the length of the pipe. Use c = 343 m/s for the speed of sound.
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To find the two lowest possible values for the length of the pipe, we can use the formula for the fundamental frequency of a closed-open tube:
f = (2n - 1) * v / (4L)
Where:
f = frequency
n = harmonic number
v = speed of sound
L = length of the tube
Since the pipe is open at both ends, we need to find the harmonics for which the pipe has resonant frequencies. The harmonics for an open pipe are odd multiples of the fundamental frequency. Therefore, we can write the following equations:
655 Hz = (2n - 1) * 343 m/s / (4L) ---(1)
786 Hz = (2n - 1) * 343 m/s / (4L) ---(2)
To find the lowest possible values for L, we need to consider the smallest possible value for n. Since n represents the harmonic number, the smallest possible value for n is 1.
Substituting n = 1 into equations (1) and (2), we get:
655 Hz = 343 m/s / (4L) ---(3)
786 Hz = 343 m/s / (4L) ---(4)
Now, we can solve equations (3) and (4) simultaneously to find the values of L.
Dividing equation (4) by equation (3), we get:
786 Hz / 655 Hz = (343 m/s / (4L)) / (343 m/s / (4L))
Simplifying further, we get:
1.2 ≈ L / L
Since L / L = 1, this equation is true for any value of L.
Therefore, the two lowest possible values for the length of the pipe are the same. We can't determine the specific length of the pipe without additional information.