A string,0.26 m long,vibrating in the n = 6 harmonic,excites an open pipe,0.88 m long,into second overtone resonance.The speed of sound in air is 345 m/s.The common resonant frequency of the string and the pipe,in SI units,is closest to:
A)590
B)390
C)490
D)690
E)760

Question

Quizplus · Accepted Answer

The common resonant frequency of the string and the pipe is the frequency at which both the string and the pipe vibrate with the same frequency. The frequency of the string is given by:f = (n * v) / 2Lwhere n is the harmonic number, v is the speed of the wave, and L is the length of the string. The frequency of the pipe is given by:f = (n + 1/2) * v / 4Lwhere n is the overtone number, v is the speed of the wave, and L is the length of the pipe. Setting these two equations equal to each other, we get:(n * v) / 2L = (n + 1/2) * v / 4LSolving for n, we get:n = 2Therefore, the common resonant frequency of the string and the pipe is the second harmonic of the string, which is:f = (2 * 345 m/s) / (2 * 0.26 m) = 590 HzTherefore, the correct answer is A.

A String,0