Question

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

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

Answer 1

The correct option is B $\lambda =\pi A/2$

In a plane progressive transverse wave particles of the medium oscillate simple harmonically about their mean positions. Hence, the concepts of SHM are applicable to the particles.
Hence, the maximum particle velocity $=\pm A\omega$ which is at the mean position.
From the given equation of wave:
$y=A\sin (2\pi ft-2\pi \frac{x}{\lambda })$

From the wave equation, $\omega =2\pi f$

and wave velocity, $v=\frac{\omega }{k}=\frac{2\pi f}{\frac{2\pi }{\lambda }}v=\lambda f$

From the given condition :
$A\times 2\pi f=4\lambda f\lambda =\frac{\pi A}{2}$