A string that is 0.26 m long is vibrating in its n = 6 harmonic. The sound from tthis string excites a pipe that is $0.88 \mathrm {~m}$ long and open at both ends into its second overtone resonance. What is the common resonant frequency of the string and the pipe? The speed of sound in air is 345 m/s.
A)590 Hz
B)390 Hz
C)490 Hz
D)690 Hz
E)760 Hz

Question

Quizplus · Accepted Answer

The string's frequency is given by:$$f_s = \frac{nv}{2L}$$where n is the harmonic number, v is the speed of the wave, and L is the length of the string.The pipe's frequency is given by:$$f_p = \frac{(2n-1)v}{4L}$$where n is the overtone number, v is the speed of the wave, and L is the length of the pipe.Setting the two frequencies equal, we get:$$\frac{nv}{2L} = \frac{(2n-1)v}{4L}$$Solving for n, we get:$$n = 6$$Substituting this value of n into either of the frequency equations, we get:$$f = \frac{6v}{2L} = \frac{6(345 \mathrm{~m/s})}{2(0.88 \mathrm{~m})} = 590 \mathrm{~Hz}$$

A String That Is 0 0.88 m0.88 \mathrm {~m}0.88 m Long and Open at Both Ends into Its Second Overtone

A String That Is 0 $0.88 \mathrm {~m}$ Long and Open at Both Ends into Its Second Overtone