Physics - Wave on String Question with Solution | TestHub

The given wave equation is $Y=Y_0\sin 2\pi (ft-x/\lambda )$.

Comparing this with the standard wave equation $Y=A\sin (\omega t-kx)$, we have amplitude $A=Y_0$, angular frequency $\omega =2\pi f$, and wave number $k=2\pi /\lambda$.

The maximum particle velocity is $v_{p,max}=A\omega =Y_0(2\pi f)$.

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

Given $v_{p,max}=4v_w$:

$Y_0(2\pi f)=4(f\lambda )$

$2\pi Y_{0}f=4f\lambda$

Dividing by $f$ (assuming $f\ne 0$):

$2\pi Y_0=4\lambda$

$\lambda =\frac{2\pi Y_0}{4}$

$\lambda =\frac{\pi Y_0}{2}$