Q. Two waves $y_{1} = A_{1}\sin{\left( \omega t - \beta_{1} \right),\ y_{2} = A_{2}\sin{(\omega t - \beta_{2})}}$ Superimpose to form a resultant wave whose amplitude is
A) $\sqrt{A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}\cos{(\beta_{1} - \beta_{2})}}$
B) $\sqrt{A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}\sin{(\beta_{1} - \beta_{2})}}$
C) $A_{1} + A_{2}$
D) $|A_{1} + A_{2}|$
Explanation
(a) Phase difference between the two waves is
$\phi = \left( \omega t - \beta_{2} \right) - \left( \omega t - \beta_{1} \right) = (\beta_{1} - \beta_{2})$
$\therefore$ Resultant amplitude $A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}\cos{(\beta_{1} - \beta_{2})}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.