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(A) For a simple harmonic oscillator,the potential energy is given by $U = \frac{1}{2}kx^2$ and the kinetic energy is given by $K = \frac{1}{2}k(A^2 -

Vedclass · Accepted Answer

$x = \pm \frac{A}{\sqrt{2}}, K = U = \frac{1}{4}kA^2$

Vedclass · Answer

(A) For a simple harmonic oscillator,the potential energy is given by $U = \frac{1}{2}kx^2$ and the kinetic energy is given by $K = \frac{1}{2}k(A^2 - x^2)$,where $A$ is the amplitude and $k$ is the force constant.At the point of intersection,$K = U$.Therefore,$\frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2)$.Canceling $\frac{1}{2}k$ from both sides,we get $x^2 = A^2 - x^2$.$2x^2 = A^2$,which implies $x^2 = \frac{A^2}{2}$.Thus,$x = \pm \frac{A}{\sqrt{2}}$.Substituting $x^2 = \frac{A^2}{2}$ into the expression for potential energy,we get $U = \frac{1}{2}k(\frac{A^2}{2}) = \frac{1}{4}kA^2$.Since $K = U$,the kinetic energy is also $\frac{1}{4}kA^2$.Thus,the coordinates of the points of intersection are $(x, E) = (\pm \frac{A}{\sqrt{2}}, \frac{1}{4}kA^2)$.

Write the coordinates of the points of intersection of the graph of kinetic energy $(K)$ and potential energy $(U)$ for a simple harmonic oscillator.

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