A particle performs linear S.H.M. When the di | vedclass

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(B) The velocity of a particle in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$.Substituting the given values:For $x_1 = 3 \ cm$,$v_1 = 8 \ cm/s$

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$\pi \ s$

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(B) The velocity of a particle in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$.Substituting the given values:For $x_1 = 3 \ cm$,$v_1 = 8 \ cm/s$: $8 = \omega \sqrt{A^2 - 3^2} \implies 8 = \omega \sqrt{A^2 - 9}$ ... $(i)$For $x_2 = 4 \ cm$,$v_2 = 6 \ cm/s$: $6 = \omega \sqrt{A^2 - 4^2} \implies 6 = \omega \sqrt{A^2 - 16}$ ... (ii)Dividing equation $(i)$ by (ii):$\frac{8}{6} = \frac{\sqrt{A^2 - 9}}{\sqrt{A^2 - 16}} \implies \frac{4}{3} = \frac{\sqrt{A^2 - 9}}{\sqrt{A^2 - 16}}$Squaring both sides:$\frac{16}{9} = \frac{A^2 - 9}{A^2 - 16} \implies 16(A^2 - 16) = 9(A^2 - 9)$$16A^2 - 256 = 9A^2 - 81 \implies 7A^2 = 175 \implies A^2 = 25 \ cm^2$.Substituting $A^2 = 25$ into equation (ii):$6 = \omega \sqrt{25 - 16} = \omega \sqrt{9} = 3\omega \implies \omega = 2 \ rad/s$.The periodic time $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \ s$.

$A$ particle performs linear $S.H.M.$ When the displacement of the particle from the mean position is $3 \ cm$ and $4 \ cm$,the corresponding velocities are $8 \ cm/s$ and $6 \ cm/s$ respectively. Its periodic time is

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