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NEET PHYSICSOscillationsMedium

Question

A point performs simple harmonic oscillation of period TT and the equation of motion is given by x=asin⁡(ωt+π/6)x = a \sin(\omega t + \pi/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?

A

T/8

B

T/6

C

T/3

D

T/12

Step-by-Step Solution

  1. Identify Velocity Equation: The displacement is x=asin⁡(ωt+π/6)x = a \sin(\omega t + \pi/6). Velocity vv is the time derivative of displacement (v=dx/dtv = dx/dt). v=ddt[asin⁡(ωt+π/6)]=aωcos⁡(ωt+π/6)v = \frac{d}{dt}[a \sin(\omega t + \pi/6)] = a\omega \cos(\omega t + \pi/6)
  2. Identify Maximum Velocity: The maximum value of cos⁡(θ)\cos(\theta) is 1. Therefore, vmax=aωv_{max} = a\omega .
  3. Set Condition: We need the time tt when v=vmax2v = \frac{v_{max}}{2}. aωcos⁡(ωt+π/6)=aω2a\omega \cos(\omega t + \pi/6) = \frac{a\omega}{2} cos⁡(ωt+π/6)=12\cos(\omega t + \pi/6) = \frac{1}{2}
  4. Solve for Time: The general solution for cos⁡θ=1/2\cos \theta = 1/2 is θ=π/3\theta = \pi/3 (for the first positive instance). ωt+π6=π3\omega t + \frac{\pi}{6} = \frac{\pi}{3} ωt=π3−π6=π6\omega t = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} Substitute ω=2πT\omega = \frac{2\pi}{T}: 2πTt=π6\frac{2\pi}{T} t = \frac{\pi}{6} t=T12t = \frac{T}{12}

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.

PHYSICSOscillationsperformssimpleharmonicoscillationperiod

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