Question

If $y_1=5\left(mm\right)]sin\pi t$ and that of $S_2$ is $y_2=5\left(mm\right)\sin (\pi t+\pi /6)$. A wave from $S_1$ reaches point A in $1sec$ while a wave from $S_2$ reaches point A in $0.5s$. The resulting amplitude at point $A$ is :-

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

Answer 1

The correct option is A $5\sqrt{2+\sqrt{3}}mm$

$Y_1=5\left(mm\right)\sin \pi t$

$Y_2=5\left(mm\right)\sin (\pi t+\frac{\pi }{6})$

from both transverse wave equation

we can get

$A_1=5mm$

$A_2=5mm$

$\varphi =\left(\pi t\right)+\pi /6-\pi t)=\frac{\pi }{6}$

where, $A_1$ and $A_2$ are amplitude of wave equation and $\varphi$ is the phase difference

resultant amplitude$=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$

$=5\sqrt{1+1+2\times 1\times 1\times \frac{\sqrt{3}}{2}}$

$=5\sqrt{2+\sqrt{3}}$

Resultant Amplitude $=5\sqrt{2+\sqrt{3}}mm$

Correct option is $\left(A\right)$