Physics - Wave on String Question with Solution | TestHub

The given equation of the transverse wave is $y=y_0\sin 2\pi \left(ft-\frac{x}{\lambda }\right)$.

The particle velocity is given by $v_p=\frac{\partial y}{\partial t}=y_0(2\pi f)\cos 2\pi \left(ft-\frac{x}{\lambda }\right)$.

The maximum particle velocity is $v_{p,\text{max}}=2\pi f{y}_0$.

The wave velocity is $v_w=f\lambda$.

According to the given condition, the maximum particle velocity is four times the wave velocity:

$v_{p,\text{max}}=4v_w$

Substituting the values, we get:

$2\pi f{y}_0=4f\lambda$

$\lambda =\frac{2\pi y_0}{4}=\frac{\pi y_0}{2}$

Answer: $\lambda =\frac{\pi y_0}{2}$