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:$$\lambda = 2L = 2(0.82 \mathrm{~m}) = 1.64 \mathrm{~m}$$The frequency of the sound wave is:$$f = v/\lambda = 345 \mathrm{~m/s} / 1.64 \mathrm{~m} = 210 \mathrm{~Hz}$$The wavelength of the standing wave on the string is:$$\lambda_s = 2L/n = 2(0.15 \mathrm{~m})/5 = 0.06 \mathrm{~m}$$The distance between a node and an adjacent antinode is:$$\lambda_s/4 = 0.06 \mathrm{~m} / 4 = 0.015 \mathrm{~m} = 15 \mathrm{~mm}$$

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