Question
The pattern of standing waves formed on a stretched string at two instants of time are shown in The velocity of two waves superimposing to form stationary waves is $360ms^{–1}$ and their frequencies are $256Hz$
  1. Calculate the time at which the second curve is plotted.
  2. Mark nodes and antinodes on the curve.
  3. Calculate the distance between A′ and C′.

Answer

Given frequency of the wave v = 256Hz$\therefore\text{T}=\frac{1}{\text{v}}=\frac{1}{256}$ second = 0.00390
$\text{T}=3.9\times10^{-3}$ seconds.
(a) In stationary wave a particle passes though it's mean position after ever $\frac{\text{T}}{4}$ time$\therefore$ in II nd curve displacement of all medium particle, are zero so
$\text{t}=\frac{\text{T}}{4}=\frac{3.9\times10^{-3}}{4}=.975\times10^{-3}\sec$
$\text{t}=9.8\times10^{-4}$ secound.
(b) Point does not vibrate i.t., their displacement is zero always so nodes A, B, C, D and E. the point A' and C' are at maximam displacement so there are anti-nodes at A' and C'. Between A' and C' $=\lambda=\frac{\text{v}}{\text{V}}=\frac{360}{256}=\frac{90}{64}=1.41\text{m}.$

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