For a particle executing S.H.M. having amplitude A, the speed of the particle is (1/3) rd of its maximum speed when the displacement from the mean position is (a) (3A)/(√(2)) (b) 2A/3 (c) (2√(2))/(3)A (d) √(2)/3A

Answer: (c). v = ω√(A² - x²), vmax = ω A √(A² - x²) = A/3 ⇒ x² = A² - A²/9 = 8A²/9 x = (2√(2))/(3)A So the answer is (c) (2√(2))/(3)A.

For a particle executing S.H.M. having amplitude A, the speed of the… | MHT CET 2024 (15 May, Shift 2)

Q20MHT CET 2024 · 15 MAY · SHIFT 2Easy

Velocity & Acceleration in SHM

For a particle executing S.H.M. having amplitude A, the speed of the particle is $\left(\frac{1}{3}\right)$ rd of its maximum speed when the displacement from the mean position is

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Answer: (c)

$v = \omega\sqrt{A^2 - x^2}$ , $v_{max} = \omega A$
$\sqrt{A^2 - x^2} = \frac{A}{3} \Rightarrow x^2 = A^2 - \frac{A^2}{9} = \frac{8A^2}{9}$
$x = \frac{2\sqrt{2}}{3}A$
So the answer is (c) $\frac{2\sqrt{2}}{3}A$ .

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