A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is cm, where

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Solution

Q.
A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is $\sqrt{α}$ cm, where $α =$ ______ . [2024]

Ans.
(12)    $v_{m e a n} = A ω \Rightarrow 10 = 4 ω$

which gives,   $ω = \frac{5}{2} r a d / s$

$v = ω \sqrt{A^{2} - x^{2}}$

$5 = \frac{5}{2} \sqrt{4^{2} - x^{2}} \Rightarrow x^{2} = 16 - 4 = 12$

   $\Rightarrow x = \sqrt{12} c m$

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