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A particle is executing SHM along a straight line. Its velocities at distances x1x_1 and x2x_2 from the mean position are v1v_1 and v2v_2, respectively. Its time period is:

A

2πx12+x22v12+v222\pi \sqrt{\frac{x_1^2 + x_2^2}{v_1^2 + v_2^2}}

B

2πx22−x12v12−v222\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}

C

2πv12+v22x12+x222\pi \sqrt{\frac{v_1^2 + v_2^2}{x_1^2 + x_2^2}}

D

2πv12−v22x12−x222\pi \sqrt{\frac{v_1^2 - v_2^2}{x_1^2 - x_2^2}}

Step-by-Step Solution

  1. Velocity-Displacement Formula: The velocity vv of a particle performing Simple Harmonic Motion at a displacement xx from the mean position is given by: v=ωA2−x2v = \omega \sqrt{A^2 - x^2} where ω\omega is the angular frequency and AA is the amplitude.
  2. Formulate Equations: Squaring the relation gives v2=ω2(A2−x2)v^2 = \omega^2 (A^2 - x^2). Applying this to the two given cases: v12=ω2(A2−x12)— (i)v_1^2 = \omega^2 (A^2 - x_1^2) \quad \text{--- (i)} v22=ω2(A2−x22)— (ii)v_2^2 = \omega^2 (A^2 - x_2^2) \quad \text{--- (ii)}
  3. Eliminate Amplitude (AA): Subtract equation (ii) from equation (i) to eliminate AA: v12−v22=ω2(A2−x12)−ω2(A2−x22)v_1^2 - v_2^2 = \omega^2 (A^2 - x_1^2) - \omega^2 (A^2 - x_2^2) v12−v22=ω2(x22−x12)v_1^2 - v_2^2 = \omega^2 (x_2^2 - x_1^2)
  4. Solve for ω\omega: ω2=v12−v22x22−x12  ⟹  ω=v12−v22x22−x12\omega^2 = \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \implies \omega = \sqrt{\frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}}
  5. Calculate Time Period (TT): The time period is T=2πωT = \frac{2\pi}{\omega}. Substituting the expression for ω\omega: T=2πx22−x12v12−v22T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}
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