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JEE Main 202120 JulyEvening Shift
Exam
JEE Main
Year
2021
Date
20 July
Shift
Evening Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2021 · 20 July · Evening Shift · Physics · Oscillations

A particle is making simple harmonic motion along the X -axis. If...

A particle is making simple harmonic motion along the XX-axis. If at a distances x1{x}_{1} and x2{x}_{2} from the mean position the velocities of the particle are v1{v}_{1} and v2{v}_{2}, respectively. The time period of its oscillation is given as:

Options

  1. AT=2πx22+x12v12−v22T = 2\pi\sqrt{\dfrac{{x}_{2}^{2} + {x}_{1}^{2}}{{v}_{1}^{2} - {v}_{2}^{2}}}
  2. BT=2πx22+x12v12+v22T = 2\pi\sqrt{\dfrac{{x}_{2}^{2} + {x}_{1}^{2}}{{v}_{1}^{2} + {v}_{2}^{2}}}
  3. CT=2πx22−x12v12+v22T = 2\pi\sqrt{\dfrac{{x}_{2}^{2} - {x}_{1}^{2}}{{v}_{1}^{2} + {v}_{2}^{2}}}
  4. DT=2πx22−x12v12−v22T = 2\pi\sqrt{\dfrac{{x}_{2}^{2} - {x}_{1}^{2}}{{v}_{1}^{2} - {v}_{2}^{2}}}

Correct answer

D. T=2πx22−x12v12−v22T = 2\pi\sqrt{\dfrac{{x}_{2}^{2} - {x}_{1}^{2}}{{v}_{1}^{2} - {v}_{2}^{2}}}

Step-by-step solution

v2=ω2(A2−x2){v}^{2} = {\omega}^{2}\left({A}^{2} - {x}^{2}\right)

A2=x12+v12ω2=x22+v22ω2{A}^{2} = {x}_{1}^{2} + {v}_{1}^{2}{\omega}^{2} = {x}_{2}^{2} + {v}_{2}^{2}{\omega}^{2}

ω2=v22−v12x12−x22{\omega}^{2} = \dfrac{{v}_{2}^{2} - {v}_{1}^{2}}{{x}_{1}^{2} - {x}_{2}^{2}}

T=2πx12−x22v22−v12T = 2\pi\sqrt{\dfrac{{x}_{1}^{2} - {x}_{2}^{2}}{{v}_{2}^{2} - {v}_{1}^{2}}}

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