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A string, 0.28 m long and vibrating in its third harmonic, excites an open pipe that is 0.82 m long into its second overtone resonance. The speed of sound in air is 345 m/s. The speed of transverse waves on the string is closest to

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Answer:

117.8 m/s

Explanation:

Given that:

String length, L = 0.28

pipe length, L' = 0.82

Speed of sound in air, v = 345 m/s

n = 3 (3rd harmonic)

Frequency, f of 3rd harmonic ;

f = (v*n) / 2L - - - - (1)

for the pipe: ; 3rd harmonic

f = (v*n) / 2L' - - - (2)

Equating (1) and (2)

(v*n) / 2L = (v*n) / 2L'

2L' * v * n = v* n * 2L

v = vL / L'

v = (345 * 0.28) / 0.82

v = 96.6 / 0.82

v = 117.80487

v = 117.8 m/s

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