The potential energy of a simple harmonic osc | vedclass

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(D) The kinetic energy at the mean position is given by the difference between the total energy and the potential energy at the mean position.$K.E._{\

Vedclass · Accepted Answer

$\pi / 100\, s$

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(D) The kinetic energy at the mean position is given by the difference between the total energy and the potential energy at the mean position.$K.E._{	ext{mean}} = E_{	ext{total}} - P.E._{	ext{mean}} = 9\, J - 5\, J = 4\, J$.At the mean position,the potential energy is $5\, J$ and the kinetic energy is maximum,so $K.E._{	ext{max}} = \frac{1}{2} m v_{	ext{max}}^2 = 4\, J$.Given $m = 2\, kg$,we have $\frac{1}{2} 	imes 2 	imes v_{	ext{max}}^2 = 4$,which implies $v_{	ext{max}}^2 = 4$,so $v_{	ext{max}} = 2\, m/s$.We know that $v_{	ext{max}} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.$A \omega = 2\, m/s$.Since $\omega = \frac{2\pi}{T}$,we have $A 	imes \frac{2\pi}{T} = 2$.Substituting $A = 0.01\, m$,we get $0.01 	imes \frac{2\pi}{T} = 2$.$T = \frac{0.01 	imes 2\pi}{2} = 0.01\pi = \frac{\pi}{100}\, s$.

The potential energy of a simple harmonic oscillator of mass $2\, kg$ in its mean position is $5\, J.$ If its total energy is $9\, J$ and its amplitude is $0.01\, m,$ its time period would be

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