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NEET PHYSICSOscillationsMedium

Question

The displacement of a particle executing simple harmonic motion is given by y=A0+Asinωt+Bcosωty = A_0 + A \sin \omega t + B \cos \omega t. Then the amplitude of its oscillation is given by:

A

A+BA + B

B

A0+A2+B2A_0 + \sqrt{A^2 + B^2}

C

A2+B2\sqrt{A^2 + B^2}

D

A02+(A+B)2\sqrt{A_0^2 + (A+B)^2}

Step-by-Step Solution

  1. Analyze the Equation: The given displacement equation is y=A0+Asinωt+Bcosωty = A_0 + A \sin \omega t + B \cos \omega t.
  2. Identify Components:
  • The term A0A_0 is a constant, representing the shift of the equilibrium position from the origin. It does not affect the amplitude of the oscillation.
  • The oscillating part consists of two simple harmonic motions: y1=Asinωty_1 = A \sin \omega t and y2=Bcosωty_2 = B \cos \omega t.
  1. Determine Phase Difference: We know that cosωt=sin(ωt+π2)\cos \omega t = \sin(\omega t + \frac{\pi}{2}). Therefore, the second wave leads the first by a phase angle of ϕ=π2\phi = \frac{\pi}{2}.
  2. Apply Superposition Principle: The resultant amplitude RR of two SHMs with amplitudes AA and BB and phase difference ϕ\phi is given by the formula : R=A2+B2+2ABcosϕR = \sqrt{A^2 + B^2 + 2AB \cos \phi}
  3. Calculate Amplitude: Substituting ϕ=π2\phi = \frac{\pi}{2}: R=A2+B2+2ABcos(90)R = \sqrt{A^2 + B^2 + 2AB \cos(90^\circ)} R=A2+B2+0R = \sqrt{A^2 + B^2 + 0} R=A2+B2R = \sqrt{A^2 + B^2}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Oscillations. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSOscillationsdisplacementparticleexecutingsimpleharmonic

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