Question

The resultant amplitude due to superposition of two waves $y_1=5\sin (wt-kx)$ and $y_2=-5\cos (wt-kx-{150}^o)$

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

Answer 1

Answer: (A)

$5$

Given :
$y_1=5\sin (wt+kx)y_2=5\cos (wt+kx+{150}^{\circ })y=y_1+y_2$
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;
$A_1=5$
$A_2=5$
$\varphi ={150}^{\circ }$
on substituting the values on the given formula:
$A=\sqrt{(5{)}^2+(5{)}^2+2\times 5\times 5\times \cos {150}^{\circ }}$
$=\sqrt{25+25-2\times 25\times \frac{1}{2}}.................(\cos {150}^{\circ }=-\frac{1}{2})$
Hence A=5 (option A is correct)