I have a question I wish to answer:
Show that the simple harmonic motion solution of the simple pendulum in the form $$\theta (t) = A\cos ({\omega _0}t)$$ (constant A) conserves net mechanical energy $E = K + U$.
I have the equation for E as $E = {1 \over 2}(m{v^2} + mgl{\theta ^2})$. I want to show its derivative is equal to 0 obviously. After I substitute in $\theta \left( t \right)$ and differentiate, I get $${{dE} \over {dt}} = {{{A^2}mgl{\omega _0}\sin (2{\omega _0}t)} \over 2}$$
Which is not 0... What am I doing wrong?
$$v=\frac{l d\theta(t)}{dt}=-l A\omega_0\sin(\omega_0t)\\E=\frac{1}{2}(mA^2l^2\omega_0^2\sin^2(\omega_0 t)+mglA^2\cos^2(\omega_0t))$$ For $\omega_0^2=g/l$ you get $E=mglA^2$. For any other values of $\omega_0\ne0$ the energy is not conserved.