A particle executes linear simple harmonic mo | vedclass

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Question

(A) Given,amplitude $A = 3\,cm$ and displacement $x = 2\,cm$.The velocity of a particle in simple harmonic motion is $v = \omega \sqrt{A^2 - x^2}$.The

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$\frac{4\pi}{\sqrt{5}}$

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(A) Given,amplitude $A = 3\,cm$ and displacement $x = 2\,cm$.The velocity of a particle in simple harmonic motion is $v = \omega \sqrt{A^2 - x^2}$.The magnitude of its acceleration is $a = \omega^2 x$.Given that $|v| = |a|$,we have $\omega \sqrt{A^2 - x^2} = \omega^2 x$.Dividing both sides by $\omega$ (assuming $\omega 
eq 0$),we get $\sqrt{A^2 - x^2} = \omega x$.Squaring both sides: $A^2 - x^2 = \omega^2 x^2$.Substituting the values: $3^2 - 2^2 = \omega^2 (2^2) \Rightarrow 9 - 4 = 4\omega^2 \Rightarrow 5 = 4\omega^2$.Thus,$\omega^2 = \frac{5}{4}$,which gives $\omega = \frac{\sqrt{5}}{2}$.The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{5}/2} = \frac{4\pi}{\sqrt{5}}\,s$.

$A$ particle executes linear simple harmonic motion with an amplitude of $3\,cm$. When the particle is at $2\,cm$ from the mean position,the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:

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