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
\([\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \sin 60^{\circ}=\cos 30^{\circ}\) \(=\sqrt{3} / 2]\)
- A 6 m
- B 12 m
- C \(12 \sqrt{3} \mathrm{~m}\)
- D \(6 \sqrt{3} \mathrm{~m}\)
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,
\(\pi =\mathrm{A} \times \frac{\pi}{6} \times \cos \left(\frac{\pi}{6} \times 2\right) \)
\( \therefore 1 =\frac{A}{6} \cos \left(\frac{\pi}{3}\right)=\frac{A}{6} \times \frac{1}{2} \)
\( \therefore A =12 \mathrm{~m}\)
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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