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

(A) $\sqrt{2} A, \frac{3 \pi}{4}$

(C) $\sqrt{3} A, \frac{5 \pi}{6}$

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Elasticity & SHM

A point mass is subjected to two simultaneous sinusoidal displacements in x-direction,

x_{1} (t) = A sin ω t

and

x_{2} (t) = A sin (ω t + \frac{2 π}{3})

. Adding a third sinusoidal displacement

x_{3} (t) = B sin (ω t + ϕ)

brings the mass to a complete rest. The values of B and

ϕ

are

Answer: (B)

Using phasor addition, the resultant of

x_{1}

and

x_{2}

is found by vector addition. The two phasors have magnitude A and are separated by

120°

. Their resultant has magnitude A (by the parallelogram law for two equal vectors at 120°) and makes an angle of

60°

with the first phasor, i.e., at phase

\frac{π}{3}

. For the mass to come to rest,

x_{3}

must cancel this resultant, so

B = A

and

ϕ = \frac{π}{3} + π = \frac{4 π}{3}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option