Question

The pattern of standing waves formed on a stretched string at two instants of time are shown in Fig. The velocity of two waves superimposing to form stationary waves is 360 ms^-1 and their frequencies are 256 Hz.

(a) Calculate the time at which the second curve is plotted.

(b) Mark nodes and antinodes on the curve.

(c) Calculate the distance between A’ and C’.

Answer 1

Given: frequency of wave, v = 256 Hz

Time period, T = \(\frac{1}{v}=\frac{1}{256}\) 

= 3.9 × 10^-3 s

(a) Time taken to pass through mean position

t = \(\frac{T}{4}\)

= \(\frac{3.9\times10^3}{4}\) s

= 9.8 × 10^-4 s

(b) Nodes- A, B, C, D, E (zero displacement)

Antinodes- A’, C’ (maximum displacement)

(c) At A’, C’, there are consecutive anti-nodes, so distance between A’ and C’,

λ = \(\frac{u}{v}=\frac{360}{256}\)

λ = 1.41 m