The maximum speed and acceleration of a particle executing simple harmonic motion are 10 $c m s^{- 1}$ and 50 $c m s^{- 2}$ . Magnitude of displacement of the particle when the speed is 2.5 $c m s^{- 1}$ is

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0.6 cm

1.2 cm

0.8 cm

$\sqrt{\frac{15}{4}} c m$

answer is D.

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Detailed Solution

$\begin{array}{l} A ω = 10; A ω^{2} = 50 \\ \Rightarrow A = 2 cm; ω = 5 {rads}^{- 1} \\ u \sin g v = ω \sqrt{A^{2} - x^{2}} \\ 2.5 = 5 \sqrt{2^{2} - x^{2}} \Rightarrow x = \pm \sqrt{\frac{15}{4}} cm . \end{array}$

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