Question

A point mass is subjected to two simultaneous sinusoidal displacements along the $x$ - direction: $x_1\left(t\right)=A\sin \omega t$ and $x_2\left(t\right)=A\sin (\omega t+\frac{2\pi }{3})$. Adding a third sinusoidal displacement $x_3\left(t\right)=B\sin (\omega t+\varphi )$ brings the mass to a complete rest. The values of $B$ and $\varphi$ are respectively:

• A. $\sqrt{2}A$ and $\frac{3\pi }{4}$
• B. $A$ and $\frac{4\pi }{3}$
• C. $\sqrt{3}A$ and $\frac{5\pi }{6}$
• D. $A$ and $\frac{\pi }{3}$

Answer 1

The correct option is B $A$ and $\frac{4\pi }{3}$

Given,
$x_1\left(t\right)=A\sin \omega t$
$x_2\left(t\right)=A\sin (\omega t+\frac{2\pi }{3})$
$x_3\left(t\right)=B\sin (\omega t+\varphi )$
We know that, mass will be at rest when $F_{net}=0$

From principle of superposition
${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3$
When the mass is at rest,
${\overrightarrow{F}}_3=-({\overrightarrow{F}}_1+{\overrightarrow{F}}_2).......\left(1\right)$

Using Hooke's law, we can rewrite $\left(1\right)$ as
$x_3=-(x_1+x_2)$
[since $F\propto x$ in SHM]
$\Rightarrow B\sin (\omega t+\varphi )=-(A\sin \omega t+A\sin (\omega t+\frac{2\pi }{3}\left)\right)$

Using, $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$
we get,

$B\sin (\omega t+\varphi )=-2A\sin (\omega t+\frac{\pi }{3})\cos \left(\frac{\pi }{3}\right)$
$\Rightarrow B\sin (\omega t+\varphi )=-A\sin (\omega t+\frac{\pi }{3})$
We know that $\sin (\pi +\theta )=-\sin \theta$
$\therefore B\sin (\omega t+\varphi )=A\sin (\omega t+\frac{4\pi }{3})$

Therfore, Amplitude of the third sinusoidal displacement is $A$ and Initial phase $\varphi =\frac{4\pi }{3}$.

Thus, option (b) is the correct answer.


Alternate solution:
To bring the mass to complete rest, the net force acting on the particle should be zero. The sinusoidal displacement vector can be represented as an equilateral triangle.



Hence, from the phasor diagram , it can be seen that amplitude of the third displacement should be $A$ and phase constant should be $\frac{4\pi }{3}$.
Hence, $\left(b\right)$ is the required answer