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 0.54 cm. The waves on the shorter string propagate with a speed of 42.3 m/s, and the fundamental frequency of the shorter string is 239 Hz. Determine the beat frequency produced by the two standing waves.

Answer 1

Assuming both pieces of string are in their fundamental frequency.

For a wave on the string, the speed v, tension T and linear density (mass per unit length, u) are related by

v = √(T/u).

You're told T and u are the same for both pieces of string, so the wave speed must be the same for both pieces of string.

Also for any wave, v = fλ, speed = frequency times wavelength. So for the shorter string,

λ = v/f

= (42.3 m/s)/(239 Hz) = 0.17699 m.

And for the fundamental wave, the length of string is half a wavelength, or

0.17699 m/2 = 0.08849m = 8.849 cm.

The longer piece of string is 0.54 cm longer than the shorter piece, so its length is (8.849 + 0.54) = 9.3894 cm, and the fundamental wavelength is double this, or

9.3894 *2 = 18.78 cm = 0.1878 m.

We have already ascertained that the speed of the waves is the same in both waves, so v = 42.3 m/s on the longer string too. And f = v/λ = (42.3 m/s)/(0.1878 m) = 225.25 Hz.

Now, the beat frequency is the difference between the two individual frequencies

= 239 Hz – 225.25 Hz

= 13.75 Hz