COMEDK · Physics · 15. Oscillations
A particle is executing simple harmonic motion. The displacement of the particle from the mean position in 2 second is equal to \(\dfrac{1}{\sqrt{2}}\) times its amplitude. What is the period of oscillation of the particle ?
- A \(\mathrm{T}=16 \mathrm{~s}\)
- B \(\mathrm{T}=4 \mathrm{~s}\)
- C \(\mathrm{T}=2 \mathrm{~s}\)
- D \(\mathrm{T}=8 \mathrm{~s}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{T}=16 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
The displacement of a particle in simple harmonic motion is given by \(x(t) = A \sin(\omega t + \phi)\). Assuming the particle starts from the mean position at \(t = 0\), we have \(\phi = 0\), so \(x(t) = A \sin(\omega t)\). Given that at \(t = 2\) s,…
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