Question

The equation of a transverse wave is given $y = 0.05\sin \pi (2t - 0.02s)$ , where $x,y$ are in meters and $t$ is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively______.
A. $50m,50m/s$
B. $100m,100m/s$
C. $50m,100m/s$
D. $100m,50m/s$

Answer 1

Compare the given equation with the standard equation of the wave form. If it is not in that form, then manipulate the given equation so that it can be compared with the standard equation. After manipulation compare the asked quantities and you will get the answer. Remember to handle the units of the equation with conscience.

Complete step-by-step solution:
The standard equation of the transverse wave is:
$y = A\sin (\omega t - kx)$
Where:
$A$ is amplitude of wave
$\omega$ is angular frequency
$t$ is time
$k$ phase difference
$x$ path difference
We have:
$k = 0.02\pi$ and $\omega = 2\pi$
Wave velocity is: $v = \dfrac{\omega }{k}$
$v = \dfrac{\omega }{k} = \dfrac{{2\pi }}{{0.02\pi }} = 100m/s$
Minimum distance of separation between two particles which are in phase is equal to the wavelength of the wave.
Wavelength: $\lambda = \dfrac{{2\pi }}{k}$
$\lambda = \dfrac{{2\pi }}{k} = \dfrac{{2\pi }}{{0.02\pi }} = 100m$
Hence, option B is correct.

Note:- Some point to remember while solving wave equation questions:
Transverse waves oscillate in the z-y plane but travel along the x axis.
A transverse wave has a speed of propagation given by the equation $v{\text{ }} = {\text{ }}f\lambda .$
The direction of energy transfer is perpendicular to the motion of the wave.