Q115 / 150 · MHT CET 2022 (05 Aug Shift 1) · Physics

For a particle performing S.H.M. the equation $\left(\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}\right)+\alpha \mathrm{x}=0$. Then the time period of the motion will be

  • A
    $\frac{2 \pi}{\alpha}$
  • B
    $2 \pi \alpha$
  • C
    $2 \pi \sqrt{\alpha}$
  • Correct answer
    D
    $\frac{2 \pi}{\sqrt{\alpha}}$

Solution

The correct option is (D). Concept: For SHM $\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}=-\left(\omega^2\right) \mathrm{x}$ is the necessary condition, where $\omega$ is the angular frequency. On comparing with the given equation $\omega=\sqrt{\alpha}$ Therefore, the time period of the SHM is: $\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\alpha}}$

Hence, the correct answer is option: 'D'.

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