The pattern of standing waves formed on a stretched string at two instants of time are shown in the figure. The velocity of two waves superimposing to form stationary waves is 360 m/s 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:

According to the quetion, the frequency of the wave is

v = 256 Hz

We know that tge time period and the frequency are related by the followsing expression:

T = 1/v

T = 1/256 sec

T = 0.00390 sec

T = 3.9 × 10^-3 sec

a) The time is computed as follows when the particle’s mean position changes after T/4:

t = T/4

t = (3.9 10^-3)/4

t = 0.975 × 10^-3 sec

t = 9.8 × 10^-4 sec

b) The nodes are given by  A, B, C, D, E

And the antinodes by  A’, C’

c) The distance between A’ and C’ is calculated as follows:

λ = v/f

λ = 360/256

λ = 1.41 m