Question

A point mass is subjected to two simultaneous sinusoidal displacements in 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.

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

Answer 1

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

The point mass was in motion due to $x_1$ and $x_2$ and the motion stopped due to $x_3$.

$\therefore \stackrel{\bullet }{x_1}+\stackrel{\bullet }{x_2}=\stackrel{\bullet }{x_3}$

$x_1=A\sin \omega t;\stackrel{\bullet }{x_1}=A\omega \cos \omega t$

$x_2=A\sin (\omega t+\frac{2\pi }{3});\stackrel{\bullet }{x_2}=A\omega \cos (\omega t+\frac{2\pi }{3})$

$x_3=B\sin (\omega t+\varphi );\stackrel{\bullet }{x_3}=B\omega \cos (\omega t+\varphi )$

$A\omega \cos \omega t+A\omega \cos (\omega t+\frac{2\pi }{3})=B\omega \cos (\omega t+\varphi )$

$⟹A[\cos \omega t+\cos \omega t\cos \frac{2\pi }{3}-\sin \omega t\sin \frac{2\pi }{3}]=B\cos (\omega t+\varphi )$

$⟹A[\cos \omega t-\frac{1}{2}\cos \omega t-\frac{\sqrt{3}}{2}\sin \omega t]=B\cos (\omega t+\varphi )$

$⟹A[\frac{1}{2}\cos \omega t-\frac{\sqrt{3}}{2}\sin \omega t]=B\cos (\omega t+\varphi )$

$⟹A[\cos \frac{4\pi }{3}\cos \omega t-\sin \frac{4\pi }{3}\sin \omega t]=B\cos (\omega t+\varphi )$

$⟹A\cos (\omega t+\frac{4\pi }{3})=B\cos (\omega t+\varphi )$

$\therefore B=A$

$\varphi =\frac{4\pi }{3}$