Question

Two waves are passing through a region in the same direction and at the same time. If the equations of these waves are,
$y_1=a\sin \frac{2\pi }{\lambda }(vt-x)$, and

$y_2=b\sin \frac{2\pi }{\lambda }\left[\right(vt-x)+x_0]$.

The amplitude of the resulting wave for $x_0=\left(\lambda /2\right)$ is,

• A. $|a-b|$
• B. $a+b$
• C. $\sqrt{a^2+b^2}$
  
• D. $\sqrt{a^2-b^2}$
  

Answer 1

The correct option is A $|a-b|$

Let ${\varphi }_1$ and ${\varphi }_2$ represent phase-angles of the first and second waves respectively, then,

${\varphi }_2=\frac{2\pi }{\lambda }\left[\right(vt-x)+x_0]$, and

${\varphi }_1=\frac{2\pi }{\lambda }(vt-x)$

Therefore, the phase difference between the waves is given by,
$\Delta \varphi ={\varphi }_2-{\varphi }_1=\frac{2\pi }{\lambda }{x}_0$
Here, $x_0=\frac{\lambda }{2}$
$\therefore \Delta \varphi =\pi$

Now, the amplitude of the resultant wave is given by,

$A=\sqrt{a^2+b^2+2ab\cos \Delta \varphi }$

$A=\sqrt{a^2+b^2+2ab\cos \pi }$

$A=\sqrt{(a-b{)}^2}$ $(\because \cos \pi =-1)$

$\therefore A=|a-b|$

Hence, $\left(A\right)$ is the correct answer.