The maximum velocity of a particle performing S.H.M. is ' V '. If the periodic time is made (1/3) rd and the amplitude is doubled, then the new maximum velocity of the particle will be (a) V/6 (b) 3V/2 (c) 3V (d) 6V

Answer: (d). vmax = (2π A)/(T), so vmax ∝ A/T A' = 2A, T' = T/3: v' = 2× 3 V = 6V So the answer is (d) 6V.

The maximum velocity of a particle performing S.H.M. is ' V '. If the… | MHT CET 2023 (9 May, Shift 1)

Q41MHT CET 2023 · 9 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM

The maximum velocity of a particle performing S.H.M. is ' $V$ '. If the periodic time is made $\left(\frac{1}{3}\right)$ rd and the amplitude is doubled, then the new maximum velocity of the particle will be

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Answer: (d)

$v_{max} = \frac{2\pi A}{T}$ , so $v_{max} \propto \frac{A}{T}$
$A' = 2A$ , $T' = \frac{T}{3}$ : $v' = 2\times 3\,V = 6V$
So the answer is (d) $6V$ .

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