Q 11-13-097JEE MainJEE Main 2022 (27 Jun, Shift 2)Easy
The equation of a particle executing simple harmonic motion is given by $x = \sin\pi\left(t + \dfrac13\right)$ m. At $t = 1$ s, the speed of particle will be (Given: $\pi = 3.14$)
Answer: (A) $157\ \text{cm s}^{-1}$
$$v = \frac{dx}{dt} = \pi\cos\pi\left(t + \frac13\right)$$
At $t = 1$ s: $v = \pi\cos\dfrac{4\pi}{3} = -\dfrac{\pi}{2}$.
Speed $= \dfrac{3.14}{2} = 1.57\ \text{m s}^{-1} = 157\ \text{cm s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics