Question

The figure below shows the snapshot of two waves $A$ and $B$ at a time instant $t$. The equation for $A$ is,
$y=A\sin (kx-\omega t-\varphi )$
and for $B$ is,
$y=A\sin (kx-\omega t)$.
It is clearly shown in the figure that wave $A$ is ahead of $B$ by a distance $\frac{\varphi }{k}$.

The motion of a single point in time, i.e. $y$ versus $t$ for two waves is best represented by,

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The equation for wave $A$ can be rewritten as :
$y=A\sin [kx-\omega t-\varphi ]$
$=A\sin \left[k\right(x-\varphi /k)-\omega t]$

$\therefore y=A\sin [kx-\omega (t+\frac{\varphi }{k})]$

While equation of wave B is $y=A\sin (kx-\omega t)$.

Comparing above equations, we can conclude that $A$ is ahead of $B$ by a distance $\Delta x=\varphi /k$
Now, phase difference between two waves is given by,

$\Delta \varphi =\frac{2\pi }{\lambda }\times \Delta x=\frac{2\pi }{T}\times \Delta t$
$\Rightarrow \frac{\Delta x}{\lambda }=\frac{\Delta t}{T}$
$\therefore \Delta t=\frac{T}{\lambda }\times \Delta x$

$=\frac{2\pi }{\omega }\times \frac{1}{\lambda }\times \frac{\varphi }{k}$

$=\frac{2\pi }{\omega }\times \frac{1}{\lambda }\times \frac{\varphi \lambda }{2\pi }=\frac{\varphi }{\omega }$

So, wave $A$ is ahead of $B$ by a time difference of $\varphi /\omega$.

Hence, $\left(B\right)$ is the correct answer.

Remember! In $y$ versus $t$, "ahead of means to the left of ", while in $y$ versus $x$ "ahead of means to the right of", if the wave travels in positive x-direction and vice versa.