Question

Two identical travelling waves, moving in the same direction, are out of phase by $\frac{\pi }{2}\text{rad}$. If amplitude of each wave is $A$, then find the resultant amplitude of the wave after superimposition.

• A. $1.41A$
• B. $2.16A$
• C. $1.6A$
• D. $6A$

Answer 1

The correct option is A $1.41A$

Let equation of wave $-1$ be given by

$x_1=A\sin (kx-\omega t)$

Equation of wave $-2$ is given by

$x_2=A\sin (kx-\omega t+\frac{\pi }{2})$

From the principle of superposition,
$x=x_1+x_2$

$\Rightarrow x=A\sin (kx-\omega t)+A\sin (kx-\omega t+\frac{\pi }{2})$

Using, $\sin C+\sin D=2\sin (\frac{C+D}{2})\cos (\frac{C-D}{2})$

$x=2A\cos \left(\frac{\pi }{4}\right)\sin (kx-\omega t+\frac{\pi }{4})$

Amplitude of the resultant wave,
$A_r=2A\cos \left(\frac{\pi }{4}\right)$

$A_r=2A\times \frac{1}{\sqrt{2}}=1.41A$

Hence, option (a) is the correct answer.