Question

In a sine wave, position of different particles at time $t=0$ is shown in figure. The equation for this wave travelling along positive $x$ -axis can be

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

Answer 1

Answer: (D)

As, it is clear from figure, at $t=0,x=0$, displacement $y=0$.
Therefore, options (a) or (d) may be correct. In case of (d);
$y=A\sin (kx-\omega t)$
$\frac{dy}{dt}=A\cos (kx-\omega t)[-\omega ]$
(after differentiating w.r.t. $t$ )
and $\frac{dy}{dx}=A\cos (kx-\omega t)[k]$
(after differentiating w.r.t. $x$ )
$\therefore \frac{\frac{dy}{dt}}{dy/dx}=\frac{-\omega A\cos (kx-\omega t)}{kA\cos (kx-\omega t)}=-\frac{\omega }{k}=-v$
$\therefore \frac{dy}{dt}=-v\left(\frac{dy}{dx}\right)$
i.e., particle velocity $=-$ (wave speed) $\times$ slope
And slope at $x=0$ and $t=0$ is positive, in figure.
Therefore, particle velocity is in negative $y$ - direction.