Question

Two waves have equations $x_1=a\sin (kx-\omega t+{\varphi }_1)$ and $x_2=a\sin (kx-\omega t+{\varphi }_2)$. If in the resultant wave the frequency and amplitude remain equal to amplitude and frequency of each superimposing waves, the phase difference between them is

• A. $\frac{\pi }{6}$
• B. $\frac{2\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is B $\frac{2\pi }{3}$

Given,
$x_1=a\sin (kx-\omega t+{\varphi }_1)$
$x_2=a\sin (kx-\omega t+{\varphi }_2)$

Resultant,
$x^{\prime }=x_1+x_2$

$x^{\prime }=a\left[\sin \right(kx-\omega t+{\varphi }_1)+\sin (kx-\omega t+{\varphi }_2\left)\right]$

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

$x^{\prime }=2a\sin (kx-\omega t+\frac{{\varphi }_1+{\varphi }_2}{2})\cos \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)$

From the data given in the question, we can say that,

$2a\cos \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)=a$
$\Rightarrow \cos \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)=\frac{1}{2}\Rightarrow \frac{{\varphi }_1-{\varphi }_2}{2}=\frac{\pi }{3}$
$\therefore {\varphi }_1-{\varphi }_2=\frac{2\pi }{3}$

Thus, option (b) is the correct answer.