A 0.13-m string,fixed at both ends and vibrating in its n = 4 harmonic,excites a pipe that is $0.88 \mathrm {~m}$ long and open at both ends,into its second overtone resonance.What is the speed of transverse waves on the string? The speed of sound in air is 345 m/s.
A)38 m/s
B)36 m/s
C)34 m/s
D)32 m/s
E)30 m/s

Question

Quizplus · Accepted Answer

The second overtone in an open pipe corresponds to the third harmonic, where the wavelength is $ \frac{2}{3} $ of the pipe length. Thus, the wavelength is $ \frac{2}{3} \times 0.88 \, \text{m} = 0.5867 \, \text{m} $. The frequency of the pipe is $ f = \frac{v}{\lambda} = \frac{345}{0.5867} \approx 588 \, \text{Hz} $. For the string in its 4th harmonic, the wavelength is $ \frac{2 \times 0.13}{4} = 0.065 \, \text{m} $. The speed of the wave on the string is $ v = f \times \lambda = 588 \times 0.065 \approx 38 \, \text{m/s} $.

A 013-M String,fixed at Both Ends and Vibrating in Its N