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(A) The total mechanical energy $(TE)$ is given by $TE = U_0 + \frac{1}{2} k A^2 = 300 \ J$,where $U_0$ is the potential energy at the mean position.G

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$100$

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(A) The total mechanical energy $(TE)$ is given by $TE = U_0 + \frac{1}{2} k A^2 = 300 \ J$,where $U_0$ is the potential energy at the mean position.Given $U_0 = 100 \ J$,we have $\frac{1}{2} k A^2 = 300 - 100 = 200 \ J$.The kinetic energy $(KE)$ of a harmonic oscillator is given by $KE = \frac{1}{2} k(A^2 - x^2)$.At $x = \frac{A}{\sqrt{2}}$,the kinetic energy is $KE = \frac{1}{2} k(A^2 - (\frac{A}{\sqrt{2}})^2) = \frac{1}{2} k(A^2 - \frac{A^2}{2}) = \frac{1}{2} k(\frac{A^2}{2}) = \frac{1}{4} k A^2$.Since $\frac{1}{2} k A^2 = 200 \ J$,then $\frac{1}{4} k A^2 = \frac{1}{2} 	imes (\frac{1}{2} k A^2) = \frac{1}{2} 	imes 200 = 100 \ J$.

$A$ linear harmonic oscillator has a total mechanical energy of $300 \ J$. If its potential energy at the mean position is $100 \ J$,find its kinetic energy at $x = +\frac{A}{\sqrt{2}}$. (in $J$)

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