Question

A particle is subjected to two SHMs of displacement $x_1=5\sin \left(\omega t\right)$ and $x_2=3\sin (\omega t+{60}^{\circ })$ respectively. Find out the equation of displacement of the resulting SHM.

• A. $x_{net}=7\sin (\omega t+{\tan }^{-}\left(\frac{5\sqrt{3}}{11}\right))$
• B. $x_{net}=7\sin (\omega t+{60}^{\circ }+{\tan }^{-}\left(\frac{5\sqrt{3}}{11}\right))$
• C. $x_{net}=7\sin (\omega t+{\tan }^{-}\left(\frac{3\sqrt{3}}{13}\right))$
• D. $x_{net}=7\sin (\omega t+{60}^{\circ }+{\tan }^{-}\left(\frac{3\sqrt{3}}{13}\right))$

Answer 1

The correct option is C $x_{net}=7\sin (\omega t+{\tan }^{-}\left(\frac{3\sqrt{3}}{13}\right))$

The equation of resulting SHM is given by
$x_{net}=A_{net}\sin (\text{Phase of}{A}_1+\alpha )$
Given, $x_1=5\sin \left(\omega t\right)$ and $x_2=3\sin (\omega t+{60}^{\circ })$
Also, $A_1=5,A_2=3$ and phase difference $\varphi ={60}^{\circ }$


Thus, using formula we have
$A_{net}=\sqrt{A_1^2+A_2^2+2\times A_1\times A_2\cos \varphi }$
$\Rightarrow A_{net}=\sqrt{5^2+3^2+2\times 5\times 3\times \cos {60}^{\circ }}$
$\Rightarrow A_{net}=\sqrt{25+9+30\times \frac{1}{2}}=\sqrt{49}=7$
Also, $\tan \alpha =\frac{A_2\sin \varphi }{A_1+A_2\cos \varphi }=\frac{A_2\sin {60}^{\circ }}{A_1+A_2\cos {60}^{\circ }}$
$\Rightarrow \tan \alpha =\frac{3\sin {60}^{\circ }}{5+3\cos {60}^{\circ }}=\frac{\frac{3\sqrt{3}}{2}}{5+\frac{3}{2}}=\frac{3\sqrt{3}}{13}$
$\Rightarrow \alpha ={\tan }^{-}\left(\frac{3\sqrt{3}}{13}\right)$
Hence, Equation of displacement is written as
$x_{net}=7\sin (\omega t+{\tan }^{-}\left(\frac{3\sqrt{3}}{13}\right))$