Question

Adjoining figure shows the snapshot of two waves A and B at any time t. The equation for A is $y=Asin(kx-\omega t-\varphi )$, and for B it 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
$y=Asin[kx-\omega t-\varphi ]$
$=Asin\left[k\right(x-\frac{\varphi }{k}-\omega t\left)\right]$
$Asin[k-\omega (t+\frac{\varphi }{k})]$
While equation of wave B is $y=Asin(kx-\omega t)$.
Comparing above equations, we can easily conclude that A is at a distance ahead of $\frac{\varphi }{k}$ from B or wave
A is ahead of B by a time difference of $\frac{\varphi }{\omega }$. So, (b) is the correct option.