ExamBro
ExamBro
JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2022
  • Medium
  • Numerical

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 : \(\pi=3.14\) )

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(B) \(157\)

Step-by-step Solution

Detailed explanation

\(x =\sin \pi\left( t +\frac{1}{3}\right)\)

\(x =\sin \left(\pi t +\frac{\pi}{3}\right)\)

\(V =\frac{ dx }{ dt }=\cos \left(\pi t +\frac{\pi}{3}\right) \pi\)

\(=-\pi \times \frac{1}{2}=157 \,cm / s\)

Same subject
Explore more questions on app