Question

The resultant amplitude due to superposition of two waves
$Y_1=5\sin (\omega t-kx)$ and $Y_2=-5\cos (\omega t-kx-{150}^{\circ })$

• A. $5$
• B. $5\sqrt{3}$
• C. $5\sqrt{2-\sqrt{3}}$
• D. $5\sqrt{2+\sqrt{3}}$

Answer 1

The correct option is A $5$

Given:
$Y_1=5\sin (\omega t-kx)$ and $Y_2=-5\cos (\omega t-kx-{150}^{\circ })=5\sin (\omega t-kx-{240}^{\circ })$
$Y=Y_1+Y_2$
where $Y$ is the resultant wave
The amplitude of this resultant wave is given by the formula
$A=\sqrt{(A_1{)}^2+(A_2{)}^2+2A_1{A}_2\cos \varphi }$
Here,
$A_1=5,A_2=5$ and $\varphi ={240}^{\circ }$
On substituting the values on the given formula
$A=\sqrt{(5{)}^2+(5{)}^2+2\times 5\times 5\cos \left({150}^{\circ }\right)}$
$=\sqrt{25+25-2\times 25\times \frac{1}{2}}$ $(\because \cos {240}^{\circ }=-\frac{1}{2})$
Hence, $A=5$