Question

Standing waves are set up on two strings fixed at each end, as shown in the drawing. The two strings have the same tension and mass per unit length, but they differ in length by 0.58 cm. The waves on the shorter string propagate with a speed of 41.9 m/s, and the fundamental frequency of the shorter string is 225 Hz. Determine the beat frequency produced by the two standing waves.

Answer 1

The beat frequency produced by the two standing waves is 13 Hz.

The wavelength of the shorter string

The wavelength of the shorter string is calculated as follows;

$L = \frac{\lambda}{2} \\\\\lambda = 2L\\\\\lambda = \frac{v}{f} \\\\\lambda = \frac{41.9}{225} \\\\\lambda = 0.186 \ m\\\\\lambda = 18.6 \ cm\\\\L= \frac{\lambda }{2} \\\\L = \frac{18.6 \ cm}{2} = 9.3\ cm$

The length of the longer string

$L_2 = 0.58 \ cm \ + 9.3 \ cm\\\\L_2 = 9.88 \ cm \\\\\lambda _2 = 2L_2\\\\\lambda _2 = 2(9.88 \ cm)\\\\\lambda_2 = 19.76 \ cm = 0.1976 \ m$

The frequency of the longer string is calculated as follows;

$v_1 = v_2\\\\f_2 = \frac{v_2}{\lambda_2} \\\\f_2 = \frac{41.9}{0.1976} \\\\f_2 = 212 \ Hz$

Beat frequency

The beat frequency produced by the two standing waves is calculated as follows;

$F_b = 225 \ Hz \ - \ 212 \ Hz\\\\F_b = 13 \ Hz$

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