Oscillations

JEE Main level, single correct, medium

31 of 194

Question 31 of 194

The equation of a particle executing simple harmonic motion is given by x=sin⁡π(t+13)mx=\sin \pi \left(t+\dfrac{1}{3}\right)m. At t=1st=1s, the speed of particle will be (Given: π=3.14\pi =3.14)

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Answer: (A) 157cms−1157cm{s}^{-1}

Given, x=sin⁡(πt+π3)x=\sin \left(\pi t+\dfrac{\pi }{3}\right)
Velocity v=dxdt=πcos⁡(πt+π3)v=\dfrac{dx}{dt}=\pi \cos \left(\pi t+\dfrac{\pi }{3}\right)
Now, at t=1st=1s, v=πcos⁡(π+π3)v=\pi \cos \left(\pi +\dfrac{\pi }{3}\right)
v=−π2=−3.142=−1.57ms−1⇒∣v∣=157cms−1v=\dfrac{-\pi }{2}=\dfrac{-3.14}{2}=-1.57m{s}^{-1}\Rightarrow \left|v\right|=157cm{s}^{-1}

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