Question

In a transverse progressive waves of amplitude A, the maximum particle velocity is four times wave velocity. Then the wave-length of wave is:

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

Answer 1

Answer: (B)

The maximum velocity of a wave is given by $v_{\max }=A\omega$ [Given: $v_{\max }=4v$ ] $\therefore$ $4v=A\omega$ $4n\lambda =A.2\pi n$ when $n=$ frequency $\therefore$ $\lambda =\frac{A\pi }{2}$