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MHT CET · Physics · Oscillations

A particle move in S.H.M. such that its acceleration is a \(=-\mathrm{px}\), where ' \(\mathrm{x}\) ' is the displacement of particle from equilibrium position and ' \(\mathrm{p}\) ' is a constant. The period of oscillation is

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

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

For SHM the acceleration a is proportional to the displacement \(x\) and it is directed towards the mean position of the particle:
\(\mathrm{a}=-\omega^2 \mathrm{x}=-\mathrm{px}\)
where, \(\omega\) is the angular frequency of the SHM.
On comparison, \(\omega=\sqrt{\mathrm{p}}\)
Thus, the period of oscillation can be written as, \(T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{p}}\)