The displacement of a particle performing S.H.M. is given by x = 5sin (3t + 3), where x is in cm and t is in second. The maximum acceleration of the particle will be (a) 15 cm s⁻² (b) 30 cm s⁻² (c) 45 cm s⁻² (d) 90 cm s⁻²

Answer: (c). From x = 5sin (3t + 3): A = 5 cm, ω = 3 rad/s amax = Aω² = 5× 9 = 45 cm s⁻² So the answer is (c) 45 cm s⁻².

The displacement of a particle performing S.H.M. is given by x = 5sin… | MHT CET 2021 (24 Sep, Shift 2)

Q53MHT CET 2021 · 24 SEP · SHIFT 2Easy

Velocity & Acceleration in SHM

The displacement of a particle performing S.H.M. is given by $x = 5\sin(3t + 3)$ , where $x$ is in $\text{cm}$ and $t$ is in second. The maximum acceleration of the particle will be

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Answer: (c)

From $x = 5\sin(3t + 3)$ : $A = 5$ cm, $\omega = 3$ rad/s
$a_{max} = A\omega^2 = 5\times 9 = 45\ \text{cm s}^{-2}$
So the answer is (c) $45\ \text{cm s}^{-2}$ .

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