The displacement of particle in S.H.M. is x = Acos (ω t + π/6). Its speed will be maximum at time (sin 90° = 1) (a) (π)/(3ω) s (b) (π)/(2ω) s (c) (π)/(ω) s (d) (π)/(4ω) s

Answer: (a). v = -Aωsin (ω t + π/6); speed is maximum when sin (ω t + π/6) = 1. ω t + π/6 = π/2 ⇒ ω t = π/3. t = (π)/(3ω) s. So the answer is (a).

The displacement of particle in S.H.M. is x = Acos (ω t + π/6). Its… | MHT CET 2025 (25 Apr, Shift 1)

Q18MHT CET 2025 · 25 APR · SHIFT 1Easy

Periodic Motion & SHM Basics

The displacement of particle in S.H.M. is $x = A\cos(\omega t + \pi/6)$ . Its speed will be maximum at time $(\sin 90^\circ = 1)$

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Answer: (a)

$v = -A\omega\sin(\omega t + \pi/6)$ ; speed is maximum when $\sin(\omega t + \pi/6) = 1$ .
$\omega t + \frac{\pi}{6} = \frac{\pi}{2} \Rightarrow \omega t = \frac{\pi}{3}$ .
$t = \frac{\pi}{3\omega}\ \text{s}$ .
So the answer is (a).

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