Physics - Wave on String Question with Solution | TestHub

The given equation is $y=2\sin \left(\omega t-kx\right)$

The maximum particle velocity $=A\omega$

And the wave velocity $=\frac{\omega }{k}$

For wave velocity to be equal to maximum particle velocity

$\frac{\omega }{k}=A\omega$

$\Rightarrow k=\frac{1}{ A}$

Since, $k=\frac{2\pi }{\lambda }$

$\lambda =2\pi A=4\pi \mathrm{cm}$.