The equation of a particle executing simple harmonic motion is given by $x=\sin \pi\left(t+\frac{1}{3}\right) m$. At $\mathrm{t}=1 \mathrm{~s}$, the speed of particle will be (Given : $\pi=3.14$ )

(B) $157 \mathrm{~cm} \mathrm{~s}^{-1}$

(A) $0 \mathrm{~cm} \mathrm{~s}^{-1}$

(C) $272 \mathrm{~cm} \mathrm{~s}^{-1}$

(D) $314 \mathrm{~cm} \mathrm{~s}^{-1}$

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Elasticity & SHM

The equation of a particle executing simple harmonic motion is given by

x = sin π (t + \frac{1}{3}) m

. At

t = 1 s

, the speed of particle will be
(Given :

π = 3.14

)

Answer: (B)

Given

x = sin π (t + \frac{1}{3})

m. Velocity

v = \frac{d x}{d t} = π cos π (t + \frac{1}{3})

m/s. At

t = 1

v = π cos π (1 + \frac{1}{3}) = π cos (\frac{4 π}{3}) = π \times (- \frac{1}{2}) = - \frac{π}{2}

m/s. Speed

∣ v ∣ = \frac{π}{2} = \frac{3.14}{2} = 1.57

m/s = 157 cm/s.

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option