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JEE Main 202227 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
27 June
Shift
Evening Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2022 · 27 June · Evening Shift · Physics · Oscillations

The equation of a particle executing simple harmonic motion is...

The equation of a particle executing simple harmonic motion is given by x=sinπ(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)

Options

  1. A157cms−1157cm{s}^{- 1}
  2. B0cms−10cm{s}^{- 1}
  3. C272cms−1272cm{s}^{- 1}
  4. D314cms−1314cm{s}^{- 1}

Correct answer

A. 157cms−1157cm{s}^{- 1}

Step-by-step solution

Given, x=sin⁡(\pit+π3)x = \sin \left(\pit + \dfrac{\pi}{3}\right)

Velocity v=dxdt=πcos(\pit+π3)v = \dfrac{d x}{d t} = \pi cos\left(\pit + \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} ⇒ \left|v\right| = 157cm{s}^{- 1}

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