(C) The velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{A^2 - x^2}$,and the magnitude of acceleration is given by $a =

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(C) The velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{A^2 - x^2}$,and the magnitude of acceleration is given by $a = \omega^2 x$.Given that the magnitude of velocity equals the magnitude of acceleration at $x = 1 \ cm$ with amplitude $A = 2 \ cm$:$\omega^2 x = \omega \sqrt{A^2 - x^2}$$\omega (1) = \sqrt{2^2 - 1^2}$$\omega = \sqrt{4 - 1} = \sqrt{3} \ rad/s$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.Substituting the value of $\omega$:$T = \frac{2\pi}{\sqrt{3}} \ s$.

Class 11 Physics Oscillations Method to determine Time Period and Frequency for diffrent type of Object of SHM

Medium

$A$ particle executes linear simple harmonic motion with an amplitude of $2 \ cm$. When the particle is at $1 \ cm$ from the mean position,the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

A
$\frac{1}{2\pi \sqrt{3}}$
B
$2\pi \sqrt{3}$
C
$\frac{2\pi}{\sqrt{3}}$
D
$\frac{\sqrt{3}}{2\pi}$

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Method to determine Time Period and Frequency for diffrent type of Object of SHM

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$A$ particle executes linear simple harmonic motion with an amplitude of $2 \ cm$. When the particle is at $1 \ cm$ 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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The ends of a rod of length $l$ and mass $m$ are attached to two identical springs. The rod is free to rotate about its centre $O$. The rod is depressed slightly at end $A$ and released. The time period of the resulting oscillation is

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