Question

When two sound waves with a phase difference of $\frac{\pi }{2}$ and each having amplitude A and frequency $\omega$, are superimposed on each other, then the maximum amplitude and frequency of resultant wave is:

• A. $\sqrt{2}A;\omega$
• B. $\frac{A}{\sqrt{2}};\frac{\omega }{2}$
• C. $\left(\sqrt{2}\right)A;\frac{\omega }{2}$
• D. $\frac{A}{2};\omega$

Answer 1

The correct option is A $\sqrt{2}A;\omega$

Given,

Frequency of both the waves =$\omega$

Amplitude of both the waves = A

Phase difference, $\theta =\frac{\pi }{2}$

When the waves are superimposed, frequency of the wave will remain same.

amplitude will be the resultant of both the waves which is given by

$A=\sqrt{{A_1}^2+{A_2}^2+2A_1{A}_2\cos \theta }$

$\Rightarrow A=\sqrt{A^2+A^2+2A.A\left(1\right)}$

$\Rightarrow A=\sqrt{2}A$

Hence, the amplitude will be$\sqrt{2}A$ and frequency will be $\omega$

Correct answer (A)