The kinetic energy of a particle,executing si | vedclass

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Question

(A) Given: Kinetic Energy $(K.E.)$ at mean position $= 16 \ J$,Amplitude $(A)$ $= 25 \ cm = 0.25 \ m$,Mass $(m)$ $= 5.12 \ kg$.The kinetic energy at t

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$\frac{\pi}{5} \ s$

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(A) Given: Kinetic Energy $(K.E.)$ at mean position $= 16 \ J$,Amplitude $(A)$ $= 25 \ cm = 0.25 \ m$,Mass $(m)$ $= 5.12 \ kg$.The kinetic energy at the mean position is equal to the total energy $(E)$ of the particle in simple harmonic motion.The formula for total energy is $E = \frac{1}{2} m \omega^2 A^2$.Substituting the values: $16 = \frac{1}{2} 	imes 5.12 	imes \omega^2 	imes (0.25)^2$.Solving for $\omega^2$: $\omega^2 = \frac{16 	imes 2}{5.12 	imes 0.0625} = \frac{32}{0.32} = 100$.Therefore,$\omega = \sqrt{100} = 10 \ rad/s$.The period of oscillation $(T)$ is given by $T = \frac{2 \pi}{\omega}$.Substituting $\omega$: $T = \frac{2 \pi}{10} = \frac{\pi}{5} \ s$.

The kinetic energy of a particle,executing simple harmonic motion is $16 \ J$ when it is in the mean position. If the amplitude of motion is $25 \ cm$ and the mass of the particle is $5.12 \ kg$,the period of oscillation is

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