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The pattern of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is $360$ $\mathrm{ms}^{-1}$ and their frequencies are $256$ $\mathrm{Hz}$.
$(a)$ Calculate the time at which the second curve is plotted.
$(b)$ Mark nodes and antinodes on the curve.
$(c)$ Calculate the distance between $\mathrm{A}^{\prime}$ and $\mathrm{C}^{\prime}$.
Solution
Here, $\mathrm{T}=\frac{1}{f}=\frac{1}{256}=3.9 \times 10^{-3} \mathrm{~s}$
$(a)$ Time taken to pass through mid point,
$=\frac{\mathrm{T}}{4}=\frac{3.9 \times 10^{-3}}{4}=9.75 \times 10^{-4} \mathrm{~s}$
$(b)$ Nodal points (nodes) or points with zero displacement are $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}, \ldots$
Antinodal points (antinodes) or points with maximum displacement are $A^{\prime}, C^{\prime}$,
$(c)$ Distance between $\mathrm{A}^{\prime}$ and $\mathrm{C}^{\prime}=$ distance between two consecutive antinodes
$\lambda=\frac{v}{f}=\frac{360}{256}=1.41 \mathrm{~m}$
