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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Physics
Chapter
Oscillations

JEE Advanced 2011 · Paper 2 · Physics · Oscillations

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

A point mass is subjected to two simultaneous sinusoidal displacements in xx-direction, x1(t)=Asin⁡ωtx_1(t)=A \sin \omega t and x2(t)=Asin⁡(ωt+2π3)x_2(t)=A \sin \left(\omega t+\frac{2 \pi}{3}\right). Adding a third sinusoidal displacement x3(t)=Bsin⁡(ωt+ϕ)x_3(t)=B \sin (\omega t+\phi) bring the mass to a complete rest. The values of BB and ϕ\phi areA point mass is subjected to two simultaneous sinusoidal displacements in x -direction, x_1(t)=A ω t and x_2(t)=A (ω

Options

  1. A2A,3π4\sqrt{2 A}, \frac{3 \pi}{4}
  2. BA,4π3A, \frac{4 \pi}{3}
  3. C3A,5π6\sqrt{3} A, \frac{5 \pi}{6}
  4. DA,π3A, \frac{\pi}{3}

Correct answer

B. A,4π3A, \frac{4 \pi}{3}

Step-by-step solution

A point mass is subjected to two simultaneous sinusoidal displacements in x -direction, x_1(t)=A ω t and x_2(t)=A (ω
Resultant amplitude of x1andx2x_1 and x_2 is AA at angle (π3)\left(\frac{\pi}{3}\right) from A1A_1. To make resultant of x1,x2andx3x_1, x_2 and x_3 to be zero. A3A_3 should be equal to AA at angle ϕ=4π3\phi=\frac{4 \pi}{3} as shown in figure. ∴\therefore Correct answer is (b). Alternate Solution If we substitute, x_1+x_2+x_3=0 or Asin⁡ωt+Asin⁡(ωt+2π3)A \sin \omega t+A \sin \left(\omega t+\frac{2 \pi}{3}\right)+B \sin (\omega t+\phi)=0 Then, by applying simple mathematics, we can prove that B=A and  ϕ=4π3\begin{array}{rlrl} B & =A \\ \text{\text{ and } } & \phi & =\frac{4 \pi}{3} \end{array} Analysis of Question (i) Question is simple. (ii) Question can be solved by applying mathematics also. (iii) Amplitudes of two or more sine or cosine functions of same frequencyω\omega can be added by vector method.
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