Question

Which of the following expression is that of a simple harmonic progressive wave?

• A. $A\sin (\omega t-kx)$
• B. $A\sin \omega t$
• C. $A\sin \omega t\cos kx$
• D. $A\cos kx$

Answer 1

The correct option is A $A\sin (\omega t-kx)$

Let us consider a wave
originating at O as shown in figure, and is travelling along +X -axis. the instantaneous simple harmonic displacement of a particle at position O is $y=A\sin \left(\omega t\right)$

where, A is the amplitude, $\omega$ is the angular velocity.

Consider a particle P at a distance x from the particle O on its right. Let the wave travel with a velocity v from left to right. Since it takes some time for the disturbance to reach P, its displacement can be written as $y=A\sin (\omega t-\varphi )$

where, $\varphi$ is the phase lag of the particle. This phase lag goes on increasing with increase in distance from O. If $\lambda$ be the wavelength, then $\lambda =2\pi$. Thus, a path difference of $x$ corresponds to a phase difference $\frac{\varphi }{2\pi }=\frac{x}{\lambda }$

$\varphi =\frac{2\pi x}{\lambda }$

Therefore,

$y=A\sin (\omega t-\frac{2\pi x}{\lambda })$

$y=A\sin (\omega t-kx)$ .........................(since, $k=\frac{2\pi }{\lambda }$ is propagation constant)