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} $