Which of the following equations correctly represents a travelling wave having wavelength \lambda = 4.0 cm, frequency v = 100 Hz and travelling in positive x-axis direction?

y = A\sin \,\pi [(0.5\,\,cm^ - 1)x - (200\,\,s^ - 1)t]

y = A\sin [(0.50\,\pi \,cm^ - 1)x - (100\,\pi \,s^ - 1)t]

y = A\sin \,\,2\pi [(0.25\,\,cm^ - 1)x - (50\,s^ - 1)t]

y = A\sin [ ( 2\pi \over 4\,cm^ - 1 )x - ( 2\pi \over 100\,s^ - 1 )t ]

Waves PYQ — JEE Main 2022 · 30 Jun · Shift 1 · Q53 with Solution

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Correct answer: D

For a wave travelling in the positive x-direction, the standard equation is

$y = A\sin(kx - \omega t)$

where

$k = \frac{2\pi}{\lambda}, \qquad \omega = 2\pi \nu$

Here,

$k = \frac{2\pi}{\lambda} = \frac{2\pi}{4\text{ cm}} = \frac{\pi}{2}\,\text{cm}^{-1} = 0.5\pi\,\text{cm}^{-1}$

$\omega = 2\pi \nu = 2\pi(100) = 200\pi\,\text{s}^{-1}$

Thus,

$y = A\sin\left[\left(0.5\pi\,\text{cm}^{-1}\right)x - \left(200\pi\,\text{s}^{-1}\right)t\right]$

Option A: $y = A\sin [(0.50\pi\,\text{cm}^{-1})x - (100\pi\,\text{s}^{-1})t]$ Here, $\omega = 100\pi$ , so frequency would be $\nu = \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} = 50\text{ Hz}$ Incorrect.
Option B: $y = A\sin 2\pi[(0.25\,\text{cm}^{-1})x - (50\,\text{s}^{-1})t]$ Expanding: $y = A\sin[(0.5\pi\,\text{cm}^{-1})x - (100\pi\,\text{s}^{-1})t]$ Again frequency is $50\text{ Hz}$ . Incorrect.
Option C: $y = A\sin \left[\left(\frac{2\pi}{4}\,\text{cm}^{-1}\right)x - \left(\frac{2\pi}{100}\,\text{s}^{-1}\right)t\right]$ The time coefficient is $\frac{2\pi}{100}$ instead of $200\pi$ . Incorrect.
Option D: $y = A\sin \pi[(0.5\,\text{cm}^{-1})x - (200\,\text{s}^{-1})t]$ Expanding: $y = A\sin[(0.5\pi\,\text{cm}^{-1})x - (200\pi\,\text{s}^{-1})t]$ This matches exactly.

The correct option is

$\boxed{\text{D}}$