Question

A transverse wave is described by the equation $ y={{y}_{0}}\sin 2\pi \left( ft-\frac{x}{\lambda } \right). $ The maximum particle velocity is equal to four times the wave velocity, if

• A. $ \lambda =\frac{\pi {{y}_{0}}}{4} $
• B. $ \lambda =\frac{\pi {{y}_{0}}}{2} $
• C. $ \lambda =\pi {{y}_{0}} $
• D. $ \lambda =2\pi {{y}_{0}} $
• E. $ \lambda =\frac{2\pi {{y}_{0}}}{3} $

Answer 1

Answer: (B)

$ y={{y}_{0}}\sin 2\pi \left( ft-\frac{x}{\lambda } \right) $ .. (i)
For particle velocity, $ \frac{dy}{dt}=2\pi f{{y}_{0}}\cos 2\pi \left( ft-\frac{x}{\lambda } \right) $
Maximum particle velocity, $ {{\left( \frac{dy}{dt} \right)}_{\max }}=2\pi f{{y}_{0}} $
Wave velocity $ =f\lambda $
Accordingly,
$ 2\pi f{{y}_{0}}=4(f\lambda ) $
Or $ \lambda =\frac{\pi {{y}_{0}}}{2} $