Question

A wave travelling in the +ve x-direction having displacement along y-direction as 1 m, wavelength 2$\pi $m and frequency of $\dfrac{1}{\pi }$ Hz is represented by
$
  {\text{A}}{\text{. }}y = \sin \left( {x - 2t} \right) \\
  {\text{B}}{\text{. }}y = \sin \left( {2\pi x - 2\pi t} \right) \\
  {\text{C}}{\text{. }}y = \sin \left( {10\pi x - 20\pi t} \right) \\
  {\text{D}}{\text{. }}y = \sin \left( {2\pi x + 2\pi t} \right) \\
$

Answer 1

Hint: We need to use the standard equation for the wave which includes amplitude, wave vector and the frequency of the wave. Then we need to find values of these parameters from the given information. Inserting these values in the standard equation, we can obtain the required answer.

Complete step-by-step answer:
We are given a wave travelling in the positive x-direction. Its displacement along y–direction is given as 1 m. This represents the amplitude of the given wave, so we its value given as
$A = 1m$
 We are given the value of its wavelength which is given as
$\lambda = 2\pi {\text{ m}}$
We are also given the value of the frequency of the wave which is given as
$\nu = \dfrac{1}{\pi }Hz$
Now we know that the standard equation for a travelling wave is given by the following expression.
$y = A\sin \left( {kx - \omega t} \right)$
Here A is the amplitude of the wave, k represents the wave vector while $\omega $ represents the angular frequency of the wave. We already have the value of amplitude A. We know that the wave vector for a wave is given as
$k = \dfrac{{2\pi }}{\lambda }$
Inserting the known values, we get
$k = \dfrac{{2\pi }}{{2\pi }} = 1rad/m$
Also, we know the expression for the angular frequency which is given as
$\omega = 2\pi \nu $
Inserting the known values, we get
$\omega = 2\pi \times \dfrac{1}{\pi } = 2rad/s$
Now we can insert all the known values in the standard wave equation. Doing so, we obtain
$y = \sin \left( {x - 2t} \right)$
This is the required answer. Hence, the correct answer is option A.

So, the correct answer is “Option A”.

Note: It should be noted that when the wave is travelling in the positive x-direction, the equation for the wave is given as $y = A\sin \left( {kx - \omega t} \right)$. But if this wave is travelling in the negative x-direction then the wave equation is given as $y = A\sin \left( {kx + \omega t} \right)$.