Q. When two sound waves with a phase difference of $\pi/2,$ and each having amplitude $A$ and frequency $\omega,$ are superimposed on each other, then the maximum amplitude and frequency of resultant wave is
A) $\frac{A}{\sqrt{2}}:\frac{\omega}{2}$
B) $\frac{A}{\sqrt{2}}:\omega$
C) $\sqrt{2}A:\frac{\omega}{2}$
D) $\sqrt{2}A:\omega$
Explanation
(d) $A_{\max} = \sqrt{A^{2} + A^{2}} = A\sqrt{2},$ frequency will remain same $i.e.\ \omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.