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Simple Harmonic Motion question

2021 · 20 Jul · Shift 2 · Q54
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  5. /2021 · 20 Jul · Shift 2 · Q54

Simple Harmonic Motion question

2021 · 20 Jul · Shift 2 · Q54

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
A particle is making simple harmonic motion along the X-axis. If at a distances x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively. The time period of its oscillation is given as :
  1. A
    T=2πx22+x12v12−v22T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 - v_2^2}}}T=2πv12​−v22​x22​+x12​​​
  2. B
    T=2πx22+x12v12+v22T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 + v_2^2}}}T=2πv12​+v22​x22​+x12​​​
  3. C
    T=2πx22−x12v12+v22T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 + v_2^2}}}T=2πv12​+v22​x22​−x12​​​
  4. D
    T=2πx22−x12v12−v22T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}}T=2πv12​−v22​x22​−x12​​​
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