Verified Solution • 2025–26 Session moderate

QJEEM22Phf5a5: JEE Main 2022 Physics - Class 11 JEEM22Phf5a5 Solution: The equation of a particle executing simple harmonic motion is…

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

(Given : π\pi = 3.14)

Multiple Choice Options

  • A 0 cm s −- 1
  • B 157 cm s −- 1
  • C 272 cm s −- 1
  • D 314 cm s −- 1

Step-by-Step Verified Solution

x=sin⁡(πt+π3)mx = \sin \left( {\pi t + {\pi \over 3}} \right)m

⇒dxdt=πcos⁡(πt+π3)\Rightarrow {{dx} \over {dt}} = \pi \cos \left( {\pi t + {\pi \over 3}} \right)

=πcos⁡(π+π3)= \pi \cos \left( {\pi + {\pi \over 3}} \right) at t=1 st = 1\,s

=−π2= - {\pi \over 2} m/s

or ∣dxdt∣=157\left| {{{dx} \over {dt}}} \right| = 157 cm/s

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