A string that is 0.15 m long and fixed at both ends is vibrating in its n = 5 harmonic. The sound from this string excites a pipe that is $0.82 \mathrm {~m}$ long and open at both ends into its second overtone resonance. What is the distance between a node and an adjacent antinode, in the string? The speed of sound in air is 345 m/s.
A)15 mm
B)7.5 mm
C)23 mm
D)30 mm
E)150 mm

Question

Quizplus · Accepted Answer

The wavelength of the sound wave in the pipe is twice the length of the pipe, or 1.64 m. The wavelength of the standing wave on the string is equal to the wavelength of the sound wave, so the distance between a node and an adjacent antinode is 1.64 m / 5 = 0.328 m = 328 mm.

A String That Is 0 0.82 m0.82 \mathrm {~m}0.82 m Long and Open at Both Ends into Its Second Overtone

A String That Is 0 $0.82 \mathrm {~m}$ Long and Open at Both Ends into Its Second Overtone