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MHT CET · Physics · Oscillations

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

  1. A \(\frac{2 \pi}{\alpha}\)
  2. B \(2 \pi \alpha\)
  3. C \(2 \pi \sqrt{\alpha}\)
  4. D \(\frac{2 \pi}{\sqrt{\alpha}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{2 \pi}{\sqrt{\alpha}}\)

Step-by-step Solution

Detailed explanation

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:
\(\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\alpha}}\)
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