Answer
$0$
Work Step by Step
For a particular transverse standing wave on a long string, one of the antinodes is at $x=0$. The displacement $y(t)$ of the string particle at $x=0$ is shown in Fig. 16-40. From the given conditions, we can write a mathematical form for $y(t)$, which takes the form
$y(x, t)=-y_m\cos kx\sin\omega t$
Now, at $x=0.10\;m$, there is a node
Therefore,
$ kx=\frac{\pi}{2}$
or,$ k=\frac{\pi}{2x}$
or, $k=\frac{\pi}{2\times0.1}\;rad/m$
or, $\boxed{k=5\pi\;rad/m}$
From the graph: $\boxed{y_m=4\;cm=0.04\;m}$
and the time period of the wave is: $T=2\;s$
Therefore, $\omega=2\pi f=\frac{2\pi}{T}=\frac{2\pi}{2}\;rad/s$
or, $\boxed{\omega=\pi\;rad/s}$
Now, the transverse velocity of the string particle is given by
$u(x,t)=-y_m\omega \cos kx\cos\omega t$
When $x=0.20\;m$ and $t=0.50\;s$ , the transverse velocity of the string particle is
$u(0.20\;m, 0.50\;s)=- 0.04\times \pi\times\cos (5\pi\times 0.20)\cos(\pi\times 0.50)\;m/s$
or, $u(0.20\;m, 0.50\;s)=0$
Therefore, the transverse velocity of the string particle is $0$.
