Q.Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.

[!TLDR] $E_k=\frac12k(A^2-x^2)$, $E_p=\frac12kx^2$; sum $E=\frac12kA^2=\frac12m\omega^2A^2$, independent of x and t, i.e. conserved. Depends on m, $ …

Kinetic energy: using $v=\omega\sqrt{A^2-x^2}$ (Eq. 5.10), $E_k=\frac12mv^2=\frac12m\omega^2(A^2-x^2)=\frac12k(A^2-x^2)$ (Eq. 5.21, using $m\omega^2=k$). Potential energy: the restoring force is $f=-kx$; the work done AGAINST it to displace the particle from 0 to x is $E_p=\int_0^xkx\,dx=\frac12kx^2=\frac12m\omega^2x^2$ (Eq. 5.23). Total energy: $E=E_k+E_p=\frac12m\omega^2(A^2-x^2)+\frac12m\omega^2x^2=\frac12m\omega^2A^2=\frac12kA^2$ (Eq. 5.24) -- the x-dependence CANCELS exactly, so E does not depend on the particle's instantaneous position (nor, by the identical cancellation of $\sin^2+\cos^2=1$ in the time-dependent forms, does it depend on t). Since m, $\omega$ and A are all fixed for a given oscillation, E is constant throughout the motion -- i.e. conserved, continuously exchanged between kinetic and potential forms but never lost. Using $\omega=2\pi n$, $E=2\pi^2mn^2A^2 …

Derive $E_k(x)$ from $v(x)$, derive $E_p(x)$ as the work done against the restoring force, add them and sho …

Forgetting to show the x-cancellation step explicitly (simply asserting energy is conserved without deriving $E_k+E_p$ algebraically …