The velocity of a particle under going SHM at the mean position is 2ms^–1. Find the velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude

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$1 .234 {ms}^{- 1}$

$1.414 {ms}^{- 1}$

$2.732 {ms}^{- 1}$

$1 .732 {ms}^{- 1}$

answer is B.

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Detailed Solution

At $y = A / 2 \Rightarrow V = ? ω A = 2 m / \sec$
$\begin{array}{l} V = ω \sqrt{A^{2} - y^{2}} = ω \sqrt{A^{2} - \frac{A^{2}}{4}} \\ V = \frac{\sqrt{3} Aω}{2} = \frac{\sqrt{3}}{2} \times 2 = \sqrt{3} m / \sec \\ V = 1 .732 m / \sec \end{array}$

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