A string that is 0.26 m long is vibrating in its n = 6 harmonic.The sound from this 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 common resonant frequency of the string and the pipe is the frequency at which both the string and the pipe will vibrate in resonance. The frequency of the string is given by:$$f_s = \frac{nv}{2L}$$where:* $f_s$ is the frequency of the string* $n$ is the harmonic number* $v$ is the speed of the wave on the string* $L$ is the length of the stringThe frequency of the pipe is given by:$$f_p = \frac{v}{2L}$$where:* $f_p$ is the frequency of the pipe* $v$ is the speed of sound in air* $L$ is the length of the pipeSetting the two frequencies equal to each other, we get:$$\frac{nv}{2L} = \frac{v}{2L}$$Solving for $n$, we get:$$n = 1$$This means that the string is vibrating in its fundamental harmonic. The frequency of the string is then:$$f_s = \frac{v}{2L} = \frac{345 \mathrm{~m/s}}{2(0.26 \mathrm{~m})} = 669.23 \mathrm{~Hz}$$The frequency of the pipe is then:$$f_p = \frac{v}{2L} = \frac{345 \mathrm{~m/s}}{2(0.88 \mathrm{~m})} = 196.08 \mathrm{~Hz}$$The common resonant frequency of the string and the pipe is therefore 196.08 Hz, which is closest to choice A, 590 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