Question

A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, ${{x}_{1}}\left( t \right)=Asin\omega t$ and ${{x}_{2}}\left( t \right)=Asin(\omega t+\dfrac{2\pi }{3})$. Adding a third sinusoidal displacement ${{x}_{3}}\left( t \right)=Bsin(\omega t+\phi )$ brings the mass to a complete rest. The values of $B$ and $\phi$ are

& (A)\sqrt{2}A,\dfrac{3\pi }{4} \\\ & (B)A,\dfrac{4\pi }{3} \\\ & (C)\sqrt{3}A,\dfrac{5\pi }{6} \\\ & (D)A,\dfrac{\pi }{3} \\\ \end{aligned}$$

Answer 1

From the question, it is clear that we have to equate the sum of all the three sinusoidal displacements in x-direction to zero, since we are told that the mass comes to rest, when the third displacement is added to the sum of first two displacements. By proceeding this way, we can find the values of $B$ and $\phi$.
Formula used:
$1)x(t)={{x}_{1}}(t)+{{x}_{2}}(t)$
$2)x(t)+{{x}_{3}}(t)=0$

Complete answer:
Given sinusoidal displacements are
${{x}_{1}}\left( t \right)=Asin\omega t$
${{x}_{2}}\left( t \right)=Asin(\omega t+\dfrac{2\pi }{3})$
${{x}_{3}}\left( t \right)=Bsin(\omega t+\phi )$
Firstly, we are required to add ${{x}_{1}}(t)$ and ${{x}_{2}}(t)$since both of these are acting in the same x-direction. If $x(t)$ represents their sum, we have
$x(t)={{x}_{1}}(t)+{{x}_{2}}(t)$
Substituting the given values of displacements in the above equation, we have
$\begin{aligned} & x(t)=Asin\omega t+Asin(\omega t+\dfrac{2\pi }{3}) \\\ & x(t)=A\sin (\omega t+\dfrac{\pi }{3}) \\\ \end{aligned}$
Now, we are provided that the mass comes to rest when a third displacement ${{x}_{3}}(t)$ is added to the above sum of displacements. Therefore, we have
$x(t)+{{x}_{3}}(t)=0\Rightarrow A\sin (\omega t+\dfrac{\pi }{3})+Bsin(\omega t+\phi )=0$
On further simplification of the above expression, we have
${{x}_{3}}(t)=B\sin (\omega t+\phi )=-A\sin (\omega t+\dfrac{\pi }{3})=A\sin (\omega t+\dfrac{4\pi }{3})$
From the above expression, it is clear that $B=A$ and $\phi =\dfrac{4\pi }{3}$.

Therefore, the correct answer is option $B$.

Note:
The given mass and displacements are related by the forces acting on the mass. Hence, we can also proceed by actually taking into consideration the forces acting on the mass and their corresponding displacements. Therefore, if ${{F}_{1}},{{F}_{2}}$ and ${{F}_{3}}$ represent the corresponding forces of the given displacements, then, as provided, we can write ${{F}_{1}}+{{F}_{2}}+{{F}_{3}}=0$ and then, proceed.