The pattern of standing waves formed on a stretched string at two… - Vidyadip

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 $360\ ms^{–1}$ and their frequencies are $256\ Hz$

Calculate the time at which the second curve is plotted.
Mark nodes and antinodes on the curve.
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 $\text{A, B, C, D}$ 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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