Question

On the superposition of the two waves represented by equation $y_1=A\sin (\omega t-kx)$ and $y_2=A\sin (\omega t-kx+\frac{\pi }{4})$, the resultant amplitude of oscillation will be:

• A. $A\sqrt{2}$
• B. $A\sqrt{2+\sqrt{2}}$
• C. $A\sqrt{2+\sqrt{3}}$
• D. $A\sqrt{1+\sqrt{2}}$

Answer 1

The correct option is B $A\sqrt{2+\sqrt{2}}$

From the given wave equations,
$A_1=A,A_2=A$
Phase difference $\left(\Delta \varphi \right)=\frac{\pi }{4}$
The resultant amplitude is,
$A_r=\sqrt{A_1^2+A_2^2+2A_1{A}_2\text{cos}\Delta \varphi }$
$A_r=\sqrt{A^2+A^2+2A^2\text{cos}\frac{\pi }{4}}$
$A_r=\sqrt{2A^2+\sqrt{2}{A}^2}$
$\therefore A_r=A\sqrt{2+\sqrt{2}}$