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The maximum particle velocity in a wave motion is half the wave velocity. Then the amplitude of the wave is equal to
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Verified Answer
The correct answer is:
$\frac{\lambda}{4 \pi}$
For a wave,
$$
y=a \sin \frac{2}{\lambda}(v t-x)
$$
Differentiating Eq. (i) w.r.t. $t$, we get
$$
\frac{d y}{d t}=\frac{2 \pi v a}{\lambda} \cos \frac{2 \pi}{\lambda}(v t-x)
$$
Now, maximum velocity is obtained when
$\cos \frac{2 \pi}{\lambda}(v t-x)=1$
$\therefore \quad v_{\max }=\left(\frac{d y}{d t}\right)_{\max }=\frac{2 \pi v a}{\lambda}$
but $\quad v_{\max }=\frac{v}{2}$
$\therefore \quad \frac{v}{2}=\frac{2 \pi v a}{\lambda} \Rightarrow a=\frac{\lambda}{4 \pi}$
$$
y=a \sin \frac{2}{\lambda}(v t-x)
$$
Differentiating Eq. (i) w.r.t. $t$, we get
$$
\frac{d y}{d t}=\frac{2 \pi v a}{\lambda} \cos \frac{2 \pi}{\lambda}(v t-x)
$$
Now, maximum velocity is obtained when
$\cos \frac{2 \pi}{\lambda}(v t-x)=1$
$\therefore \quad v_{\max }=\left(\frac{d y}{d t}\right)_{\max }=\frac{2 \pi v a}{\lambda}$
but $\quad v_{\max }=\frac{v}{2}$
$\therefore \quad \frac{v}{2}=\frac{2 \pi v a}{\lambda} \Rightarrow a=\frac{\lambda}{4 \pi}$
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