A string,0.13 m long,vibrating in the n = 4 harmonic,excites an open pipe,0.88 m long,into second overtone resonance.The speed of sound in air is 345 m/s.The velocity of transverse waves in the string,in SI units,is closest to:
A)38
B)36
C)34
D)32
E)30

Question

Quizplus · Accepted Answer

The second overtone in an open pipe corresponds to the third harmonic, which means the wavelength is 2/3 of the pipe length. The wavelength in the pipe is 0.88 m * 2/3 = 0.5867 m. The frequency is speed of sound / wavelength = 345 m/s / 0.5867 m = 588.1 Hz. For the string in the 4th harmonic, the wavelength is 2 * length of string / 4 = 0.065 m. The velocity of transverse waves in the string is frequency * wavelength = 588.1 Hz * 0.065 m = 38.23 m/s, closest to 38 m/s.

A String,0