Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • 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.

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’

Answers (1)

Frequency of wave \nu=256Hz

T=\frac{1}{\nu}=\frac{1}{256}second=0.00390=3.9\times 10^{-3}seconds

  1. In stationary wave, a particle passes through its mean position after every \frac{T}{4} time

T=\frac{T}{4} =3.9\times \frac{10^{-3}}{4}=0.975\times 10^{-3} sec=9.75\times 10^{-4}sec

  1. Point does not vibrate at A, B, C,D and E. The point A’ and C’ are at maximum displacement

Between A’ and C’ = \lambda =\frac{V}{v}=\frac{360}{256}=1.41m

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question