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 second overtone resonance of an open pipe is given by:$$f = (3/2) (v/2L) = (3/4) (v/L)$$where v is the speed of sound and L is the length of the pipe.The frequency of the string is given by:$$f = n(v/2L)$$where n is the harmonic number and L is the length of the string.Setting these two equations equal to each other, we get:$$(3/4) (v/L) = n(v/2L)$$$$3/4 = 2n$$$$n = 3/8$$This is not a valid harmonic number, so there is no common resonant frequency for the string and the pipe.

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