IIT-JEE 2011 · Paper 2 · Physics Q24

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PhysicsSingle correct+3 / -1Question 24 of 60

A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, $x_1(t)=A\sin\omega t$ and $x_2(t)=A\sin\left(\omega t+\dfrac{2\pi}3\right)$ . 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

\sqrt2A

\dfrac{3\pi}4

A

\dfrac{4\pi}3

\sqrt3A

\dfrac{5\pi}6

A

\dfrac\pi3

Answer and worked solution

Official answer: B

Add the first two. Phasors of length $A$ at angles 0 and $\frac{2\pi}3$ sum to length $A$ at angle $\frac\pi3$ :
$x_1+x_2=A\sin\left(\omega t+\frac\pi3\right).$

Cancel it. $x_3=-A\sin\left(\omega t+\frac\pi3\right)=A\sin\left(\omega t+\frac{4\pi}3\right)$ , so $B=A$ and $\phi=\frac{4\pi}3$ .

(B)

Topics:Superposition of SHMs

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IIT-JEE 2011 · Paper 2, Physics question 24