Energy in simple harmonic motion
Let us consider the kinetic and potential energy for a spring-mass system. The kinetic energy is
\[E_{\mathrm{K}} = \frac{1}{2} m v^2,\]
whilst the potential energy for the spring-mass system is
\[E_{\mathrm{P}} = \frac{1}{2} k x^2,\]
which means that mechanical energy is
\[E_{\mathrm{M}} = E_{\mathrm{K}} + E_{\mathrm{P}} = \frac{1}{2} m v^2 + \frac{1}{2} k x^2.\]
Is the potential energy formula \(E_{\mathrm{P}} = \frac{1}{2} k x^2\) true for all kinds of simple harmonic motion?
The definition of simple harmonic motion is that
\[a \propto -x.\]
Since \(F = m a\) according to Newton's second law, the net force must also be proportional to the position:
\[F = -c \cdot x,\]
where \(c\) is a constant for a given system.
The potential energy is:
\[E_{\mathrm{P}}(x) = W_{x \rightarrow 0} = \int\limits_{x}^{x_0}F(x')\mathrm{d}x' = \int\limits_{x}^{x_0}\left(-c \cdot x'\right) \mathrm{d}x' = -c \left[\frac{x'^2}{2}\right]_{x}^{x_0} = \frac{1}{2} c x^2 - \frac{1}{2} c x_0^2.\]
If the reference position is in the origin, the general formula of the potential energy is identical in form to the potential energy of the spring-mass system:
\[E_{\mathrm{P}}(x) = \frac{1}{2} c x^2.\]
The table below shows the values of the different forms of energy in certain positions of the system. We can see that the law of conservation of mechanical energy is fulfilled as the mechanical energy is the same in all these positions.
The mechanical energy is
\[E_{\mathrm{M}} = E_{\mathrm{K}} + E_{\mathrm{P}} = \frac{1}{2} m v^2 + \frac{1}{2} k x^2,\]
\[E_{\mathrm{M}} = \frac{1}{2} m A^2 \omega^2 \cos^2\left(\omega t + \phi_0\right) + \frac{1}{2} k A^2 \sin^2\left(\omega t + \phi_0\right).\]
There exists a relationship between \(k\) and \(\omega\):
\[\omega^2 = \frac{k}{m} \ \Rightarrow\ m \omega^2 = k.\]
Thus the mechanical energy is a constant value during the motion:
\[E_{\mathrm{M}} = \frac{1}{2} k A^2 \cos^2\left(\omega t + \phi_0\right) + \frac{1}{2} k A^2 \sin^2\left(\omega t + \phi_0\right),\]
\[E_{\mathrm{M}} =\frac{1}{2} k A^2 \cdot \left[\cos^2\left(\omega t + \phi_0\right) + \sin^2\left(\omega t + \phi_0\right)\right] = \frac{1}{2} k A^2.\]