Question

The resultant amplitude due to superposition of two waves
$y_1=5\sin (wt-kx)$ and
$y_2=-5\cos (wt-kx-{30}^{\circ }$)

• 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 $5$

$y_1=5\sin (\omega t-kx)$
$y_2=-5\cos (\omega t-kx-{30}^{\circ })$
$y_2=5sin(\omega t-kx-{30}^{\circ }-{90}^{\circ })$

$\Rightarrow$$y_2=5\sin (\omega t-kx-{120}^0)$
From superposition, the resultant amplitude is given by
$A_{net}=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \delta }$
$=\sqrt{5^2+5^2+2\times 5\times 5\cos {120}^{\circ }}$
$=5$

Why this question? Tip: Always try to convert the equation of waves in sin function when solving question related to superposition to maintain uniformity.