Wave Mechanics Question 621
Question: A transverse wave is represented by $ y=A\sin (\omega t+kx) $ . For what value of the wavelength is the wave velocity equal to the maximum particle velocity?
Options:
A) $ \frac{\pi A}{2} $
B) $ \pi A $
C) $ 2\pi A $
D) $ A $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ y=A\sin (\omega t-kx) $ Particle velocity, $ v _{p}=\frac{dy}{dt}=A,\omega \cos (\omega t-kx) $
$ \
Therefore {v _{p,\max }}=A,\omega $ wave velocity $ =\frac{\omega }{k} $
$ \
Therefore ,A,\omega =\frac{\omega }{k} $
$ i.e.,A=\frac{1}{k} $ But $ k=\frac{2\pi }{\lambda } $
$ \
Therefore \lambda =2\pi A $
BETA
