Question

A progressive wave travelling along the positive $x$ direction is represented by $y(x,t)=A\sin (kx-\omega t+\varphi )$. Its snapshot at $t=0$ is given in the figure.


For this wave, the phase constant $\varphi$ is :

• A. $-\frac{\pi }{2}$
• B. $\pi$
• C. $0$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\pi$

Given,
$y(x,t)=A\sin (kx-\omega t+\varphi )$
At $t=0$
$y\left(x\right)=A\sin (kx+\varphi ).......\left(1\right)$


At $x=0,y=0$ , slope is negative.
So, we can see that, as $x$ increases in the positive direction, $y$ becomes negative.

Thus, $y\left(x\right)=-A\sin \left(kx\right)$

$\Rightarrow y\left(x\right)=A\sin (kx+\pi ).......\left(2\right)$

Equating Eq.(1) to Eq. (2), we get
$\varphi =\pi$

Thus, option (b) is the correct answer.