Physics - Wave on String Question with Solution | TestHub

Comparing the given wave equation with the standard wave equation $y=a\sin (kx-\omega t)$, we can identify the wave parameters.

From the equation:
Wave number, $k=12\pi \text{rad/m}$
Angular frequency, $\omega =3000\pi \text{rad/s}$

The wavelength $\lambda$ is given by:
$\lambda =\frac{2\pi }{k}=\frac{2\pi }{12\pi }=\frac{1}{6}\text{m}$

The wave velocity $v$ is given by:
$v=\frac{\omega }{k}=\frac{3000\pi }{12\pi }=250\frac{\text{m}}{\text{s}}$

Thus, $\lambda =\frac{1}{6}\text{m}$ and $v=250\frac{\text{m}}{\text{s}}$.

Answer: $\lambda =\frac{1}{6}\text{m},v=250\frac{\text{m}}{\text{s}}$