Question

A transverse wave is described by the equation $Y=Y_0\sin 2\pi (ft-\frac{x}{\lambda })$. The maximum particle velocity is equal to four times the wave velocity, if

• A. $\lambda =\frac{\pi Y_0}{4}$
• B. $\lambda =\frac{\pi Y_0}{2}$
• C. $\lambda =\pi Y_0$
• D. $\lambda =2\pi Y_0$

Answer 1

The correct option is B $\lambda =\frac{\pi Y_0}{2}$

Given that particle velocity is $4$ times wave velocity ${\left(V_p\right)}_{max}=4v$

$Y_0\omega =4\left(f\lambda \right)$

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

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