Question

If two waves represented by
$y_1=4\sin \omega t$ , and
$y_2=3\sin (\omega t+\frac{\pi }{3})$
(where $y_1,y_2$ are in $\text{cm}$ and $t$ is in $\text{sec}$),
interfere at a point, the resultant amplitude of that point will be about,

• A. $7\text{cm}$
• B. $6\text{cm}$
• C. $5\text{cm}$
• D. $3.5\text{cm}$

Answer 1

The correct option is B $6\text{cm}$

Given:
$y_1=4\sin \omega t$
$y_2=3\sin (\omega t+\frac{\pi }{3})$

The phase difference between the waves is given by,
$\Delta \varphi =\frac{\pi }{3}\text{rad}$

So, the amplitude of the resulting wave is,
$A=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos \varphi }$
$A=\sqrt{4^2+3^2+2\times 4\times 3\cos \pi /3}$
$\Rightarrow A\approx 6\text{cm}$

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