Question

Derive expressions for potential energy and kinetic energy of an oscillating spring-
mass system

Answer 1

The equation of harmonic motion:

$x = A\sin(\omega t + \phi_0)$ , where $\omega = \sqrt{\frac{k}{m}}$ in case oscillating spring-mass system.

$\phi_0$ - initial oscillation phase (actually if it is, for example, $-\frac{\pi}{4}$ we can rewrite it with cosine)

$\dot x = A \cdot \cos(\omega t + \phi_0) \cdot \omega$

$E_k = \frac{mv^2}{2} = \frac{m\dot x^2}{2} = \frac{m(A\omega \cos(\omega t + \phi_0))^2}{2} = \frac{m\omega^2(A \cos(\omega t + \phi_0))^2}{2} = \frac{m\frac{k}{m}(A \cos(\omega t + \phi_0))^2}{2} \Rightarrow$

$$E_k = \frac{k(A\cos(\omega t + \phi_0))^2}{2} = \frac{k(A\cos(\sqrt{\frac{k}{m}} t + \phi_0))^2}{2}$$

$E_p = \frac{kx^2}{2} \Rightarrow$

$$E_p = \frac{k(A\sin(\omega t + \phi_0))^2}{2} = \frac{k(A\sin(\sqrt{\frac{k}{m}} t + \phi_0))^2}{2}$$

As we can see expressions differ only in the initial phase (we can change the sine to cosine by changing the initial phase). with Pythagorean trigonometric identity:

$$E_{mechanical} = E_p + E_k = \frac{kA^2}{2} = const$$