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JEE Main 202310 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
10 April
Shift
Morning Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2023 · 10 April · Morning Shift · Physics · Oscillations

A particle executes S.H.M. of amplitude A along x -axis. At t = 0 ,...

A particle executes S.H.M. of amplitude AA along xx-axis. At t=0t = 0, the position of the particle is x=A2x = \dfrac{A}{2} and it moves along positive xx-axis. The displacement of particle in time tt is x=Asin(\omegat+δ)x = Asin\left(\omegat + \delta\right), then the value δ\delta will beA particle executes S.H.M. of amplitude A along x -axis. At t = 0 , the position of the particle is x = (A)/(2) and it

Options

  1. Aπ2\dfrac{\pi}{2}
  2. Bπ6\dfrac{\pi}{6}
  3. Cπ3\dfrac{\pi}{3}
  4. Dπ4\dfrac{\pi}{4}

Correct answer

B. π6\dfrac{\pi}{6}

Step-by-step solution

A particle executes S.H.M. of amplitude A along x -axis. At t = 0 , the position of the particle is x = (A)/(2) and it

Given, the instantaneous equation of motion of the particle is

x=Asin(\omegat+δ)...(1)x = Asin\left(\omegat + \delta\right) . . . \left(1\right)

As the particle is moving along positive xx-axis, v=dxdt>0v = \dfrac{dx}{dt} > 0.

Substitute 00 for tt and A2\dfrac{A}{2} for xx into equation (1) and solve to calculate the required phase difference.

{A2=Asin(ω×0+δ)⇒sinδ=12=sinπ6⇒δ=π6\begin{cases}\dfrac{A}{2} & = & Asin\left(\omega \times 0 + \delta\right) \\ & ⇒ & sin\delta = \dfrac{1}{2} \\ & = & sin\dfrac{\pi}{6} \\ & ⇒ & \delta = \dfrac{\pi}{6}\end{cases}

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