Question

Let a wave $y(x, t)=a \sin (k x-\omega t)$ is reflected from an open boundary and then the incident and reflected waves overlap. Then the amplitude of resultant wave is

• A. $2 a \cos k x$
• B. $2 a \sin k x$
• C. $2 a \sin \left(\frac{k x}{2}\right)$
• D. $a \sin k x$

Answer 1

Answer: (B)

We have, incident wave, $y_{1}=a \sin (k x-\omega t)$
So the reflected wave, $y_{2}=a \sin (k x+\omega t)$
From principle of superposition,
The standing wave equation is obtained as
$y(x, t) =y_{1}+y_{2}=a[\sin (k x-\omega t)+\sin (k x+\omega t)]$
$=2 a \sin k x \cos \omega t$
On comparing Eq. (i) with $y(x, t)=A(x) \cos \omega t$, we get Amplitude, $A(x)=2 a \sin k x$