Question

On the superposition of the two waves given as $y_1=A_0\sin (\omega t-kx)$ and $y_2=A_0\cos (\omega t-kx+\frac{\pi }{6})$ the resultant amplitude of oscillation will be

• A. $\sqrt{3}{A}_0$
• B. $\frac{A_0}{2}$
• C. $A_0$
• D. $\frac{3}{2}{A}_0$

Answer 1

Answer: (C)

$A_0$

Given,

$y_1=A_{0}sin(\omega t-kx)$

$y_2=A_{0}cox(\omega t-kx+\frac{\pi }{6})=A_{0}sin(\omega t-kx+2\pi /3)$

$A_1=A_0$

$A_2=A_0$

$\theta =2\pi /3$

Resultant amplitude,

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

$R=\sqrt{A_0^2+A_0^2+2A.Acos{120}^0}$

$R=\sqrt{2A_0^2-A^2}$

$R=A_0$

The resultant amplitude is $A_0$.

The correct option is C.