Wave Mechanics Question 633
Question: A transverse sinusoidal wave, moves along a string in the positive x-direction at a speed of 10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snap-shot of the wave is shown in figure. The velocity of point P when its displacement is 5 cm is
Options:
A) $ \frac{\sqrt{3\pi }}{50}\hat{j}m/s $
B) $ -\frac{\sqrt{3\pi }}{50}\hat{j}m/s $
C) $ \frac{\sqrt{3\pi }}{50}\hat{i}m/s $
D) $ -\frac{\sqrt{3\pi }}{50}\hat{i}m/s $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Since the wave is sinusoidal moving in positive x-axis the point will move parallel to y-axis
Therefore options [c] and [d] are ruled out. As the wave moves forward in positive X-direction, the point should move upwards i.e. in the positive Y-direction.
Therefore correct option is a. Alternate solution-1: Equation of a wave moving in positive x-axis is given as $ y=A\sin (\omega t-\phi ) $ or $ v _{p}=A\omega ,\cos ,(\omega t-\phi ) $ Here y= 5 cm, A= 10 cm,
$ \
Therefore 5=10\sin (\omega t-\phi )\Rightarrow \omega t-\phi =30{}^\circ $ Substituting this value in the equation of velocity we get $ v _{p}=0.10\times \omega ,\cos 30{}^\circ $ Now $ v=v\lambda ,\
Therefore ,v=\frac{v}{\lambda }=\frac{0.10}{0.5}=0.2 $
$ \
Therefore ,\omega =2\pi v=2\pi \times 0.2=0.4\pi $
$ \Rightarrow ,v _{p}=0.1\times 0.4\pi \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{50}\pi $ It has to be in positive y direction.
BETA
