Question

A transverse wave is described by the equation $y=y_o\sin 2\pi (ft-x/a)$. The maximum particle velocity is equal to four times the wave velocity if a is equal to :

• A. $\pi y_o/4$
• B. $\pi y_o/2$
• C. $\pi y_o$
• D. $2\pi y_o$

Answer 1

The correct option is B $\pi y_o/2$

$y=y_0\sin 2\pi (ft-\frac{x}{\alpha })$

We know standard equation of wave:

$y=r\sin 2\pi (vt-\frac{x}{\lambda })$

By comparison we get,

$r=y_0,v=f,\lambda =a$

We know,

Wave velocity , $v=\frac{\omega }{k}$

And, particle velocity, $v_p=y_0\omega$

Given , $v_p=4v$

$y_0\omega =4\left(\frac{\omega }{k}\right)$

$y_0=\frac{4}{k}$

We know , $k=\frac{2\pi }{\lambda }$

Thus,

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

Or, $a=\lambda =\frac{\pi }{2}{y}_0$