Question

One string of a certain musical instrument is 78.0cm long and has a mass of 8.78g . It is being played in a room where the speed of sound is 344 m/s.

Part A

To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.754m ? (Assume that the breaking stress of the wire is very large and isn

Answer 1

The frequency of the second overtone is the velocity of propagation divided by the wavelength. Because the second overtone is the third harmonic, it vibrates at three times the frequency of the fundamental and the wavelength on the string will be exactly 2/3 the length of the string. Answering part B will be easier, so we will do that first. The frequency of the second overtone will be 344 m/s / 0.757m = 454.4 Hz. The fundamental will be 1/3 this frequency, or 151.47 Hz, which turns out to be a note about midway between A flat and A natural three octaves and a sixth above middle C. The fundamental will have a wavelength twice as long as the string and the velocity will be the wavelength times the frequency. We can therefore find the velocity of propagation along the string, which is given by the formula v = ?(T/?) where v is the velocity of propagation, T is the tension in the string, and ? is the mass per unit length of the string. The tension is therefore T = v