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MHT CET · Physics · Oscillations

A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is \(\pi \mathrm{m} / \mathrm{s}\). Amplitude of the oscillation is
\([\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \sin 60^{\circ}=\cos 30^{\circ}\) \(=\sqrt{3} / 2]\)

  1. A 6 m
  2. B 12 m
  3. C \(12 \sqrt{3} \mathrm{~m}\)
  4. D \(6 \sqrt{3} \mathrm{~m}\)
Verified Solution

Answer & Solution

Correct Answer

(B) 12 m

Step-by-step Solution

Detailed explanation

Displacement of the particle, \(x=A \sin \omega t\)
Velocity of the particle,
\(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{A} \omega \cos \omega \mathrm{t}...(i)\)
Given that,
\(\mathrm{v}=\pi \mathrm{m} / \mathrm{s}, \mathrm{~T}=12 \mathrm{~s},\)
\(\therefore \omega=\frac{2 \pi}{\mathrm{~T}}=\frac{\pi}{6} \mathrm{rad} / \mathrm{s}\)
Substituting in equation (i), we get,
\(\pi =\mathrm{A} \times \frac{\pi}{6} \times \cos \left(\frac{\pi}{6} \times 2\right) \)
\( \therefore 1 =\frac{A}{6} \cos \left(\frac{\pi}{3}\right)=\frac{A}{6} \times \frac{1}{2} \)
\( \therefore A =12 \mathrm{~m}\)
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