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 : \(\pi=3.14\) )
- A \(0\)
- B \(157\)
- C \(272\)
- D \(314\)
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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