Answer
$0.04$
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, $\boxed{\omega=\pi\;rad/s}$
When $t=0.50\;s$, and $x=0.20\;m$, the displacement of the string particle is
$y(0.20\;m, 0.50\;s)=- 0.04\cos (5\pi\times 0.20)\sin(\pi\times 0.50)\;m$
or, $y(0.20\;m, 0.50\;s)=0.04\;m$
Therefore, the displacement is $0.04$.
