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 0.51 cm. The waves on the shorter string propagate with a speed of 42.6 m/s, and the fundamental frequency of the shorter string is 232 Hz. Determine the beat frequency produced by the two standing waves.

Answer 1

Given,

The waves on the shorter string propagate with a speed, v = 42.6 m/s

The difference in length of two strings, l₂ - l₁ = 0.51 cm -------- (1)

The frequency of standing wave in stretched string is, n₁ = v / 2l1 = 232 Hz

The length of the string, l₁ = v / 2n₁

= (42.6 m/s) / 2*232Hz

= 9.18 cm

From (1) , l₂ = 0.51 + l₁

l₂ = 0.51 + 9.18 = 9.69 cm

The frequency of standing wave is longer string is, n₂ = v / 2 l₂

= (42.6 m/s) / 2 * 9.69 cm

= (42.6 m/s) / (2 * 9.69 * 10^-2)

= 219.81 Hz

The number of beats = 232 Hz - 219.81 Hz

= 12.19 Hz