The equation of a particle executing simple harmonic motion is given by x = sin π(t + 1/3) m. At t = 1s, the speed of particle will be (Given: π = 3.14) (a) 0 cm s ⁻¹ (b) 157 cm s ⁻¹ (c) 272 cm s ⁻¹ (d) 314 cm s ⁻¹

Answer: (b). v = dx/dt = πcos π(t + 1/3) m/s At t = 1 s: cos 4π/3 = -1/2, so |v| = π/2 = 1.57 m/s So the answer is (b) 157 cm s⁻¹.

The equation of a particle executing simple harmonic motion is given… | JEE Main 2022 (27 Jun, Shift 2)

Q47JEE Main 2022 · 27 JUN · SHIFT 2Easy

Velocity & Acceleration in SHM

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

(Given: $\pi$ = 3.14)

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Answer: (b)

$v = \frac{dx}{dt} = \pi\cos\pi\left(t + \frac{1}{3}\right)$ m/s
At $t = 1$ s: $\cos\frac{4\pi}{3} = -\frac{1}{2}$ , so $|v| = \frac{\pi}{2} = 1.57$ m/s
So the answer is (b) $157\ \text{cm s}^{-1}$ .

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