A particle is performing S.H.M. with an amplitude 4 cm. At the mean position the velocity of the particle is 12 cm/s. When the speed of the particle becomes 6 cm/s, the distance of the particle from mean position is (a) √(3) cm (b) √(6) cm (c) 2√(3) cm (d) 2√(6) cm

Answer: (c). vmax = Aω ⇒ ω = 12/4 = 3 rad/s v = ω√(A² - x²): 6 = 3√(16 - x²) ⇒ 16 - x² = 4 x = √(12) = 2√(3) cm So the answer is (c) 2√(3) cm.

A particle is performing S.H.M. with an amplitude 4 cm. At the mean… | MHT CET 2024 (9 May, Shift 1)

Q22MHT CET 2024 · 9 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM

A particle is performing S.H.M. with an amplitude 4 cm. At the mean position the velocity of the particle is $12\text{ cm/s}$ . When the speed of the particle becomes $6\text{ cm/s}$ , the distance of the particle from mean position is

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Answer: (c)

$v_{max} = A\omega \Rightarrow \omega = \frac{12}{4} = 3\ \text{rad/s}$
$v = \omega\sqrt{A^2 - x^2}$ : $6 = 3\sqrt{16 - x^2} \Rightarrow 16 - x^2 = 4$
$x = \sqrt{12} = 2\sqrt{3}\ \text{cm}$
So the answer is (c) $2\sqrt{3}$ cm.

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Velocity & Acceleration in SHM question from MHT CET 2024 (9 May, Shift 1)

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