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JEE Mains · Physics · STD 11 - 13. oscillations

The equation of a particle executing simple harmonic motion is given by \(x =\sin \pi\left( t +\frac{1}{3}\right) m\). At \(t =1 \,s\), the speed of particle will be .......... \(cm s ^{-1}\) (Given : \(\pi=3.14\) )

  1. A \(0\)
  2. B \(157\)
  3. C \(272\)
  4. D \(314\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(157\)

Step-by-step Solution

Detailed explanation

\(x =\sin \pi\left( t +\frac{1}{3}\right)\) \(x =\sin \left(\pi t +\frac{\pi}{3}\right)\) \(V =\frac{ dx }{ dt }=\cos \left(\pi t +\frac{\pi}{3}\right) \pi\) \(=-\pi \times \frac{1}{2}=157 \,cm / s\)
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