Question

A transverse wave is given by $y=A\sin 2\pi \left(\frac{t}{T}-\frac{x}{\lambda }\right).$ The maximum particle velocity is equal to $4$ times the wave velocity, when:

• A. $\lambda =2\pi A$
• B. $\lambda =\frac{1}{2}\pi A$
• C. $\lambda =\pi A$
• D. $\lambda =\frac{1}{4}\pi A$

Answer 1

Answer: (B)

The given equation of transverse wave is
$y=A\sin 2\pi \left(\frac{t}{T}-\frac{x}{\lambda }\right)$
Maximum particle velocity $(u_m)=a\omega$
From standard wave equation,
$a=A,\omega =2\pi n=2\pi \frac{v}{\lambda }$
$\therefore u_m=A.2\pi \frac{v}{\lambda }=4v$
$\Rightarrow \lambda =\frac{\pi A}{2}$