# Simple Harmonic Oscillation
$$F = ma = m\frac{v^2}{r} = m\omega^2r$$

$$F = kr$$

$$m\omega^2r = kr$$

$$k = m\omega^2$$
:::success
$$\omega = \sqrt{\frac{k}{m}}$$

$$T = 2\pi/\omega = 2\pi\sqrt{\frac{m}{k}}$$
:::
* Consider the mass of the spring ($m_s$):
[why ??](https://hackmd.io/bkcNhrbuRreR-OI7OkTOJQ)
$$T = 2\pi\sqrt{\frac{m+\frac{1}{3}m_s}{k}}$$
## Energy
$$U(t) = \frac{1}{2}kx^2 = \frac{1}{2}kx_m^2cos^2(\omega t+\phi)$$

$$K(t) = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2x_m^2sin^2(\omega t+\phi)$$

$$K(t) = \frac{1}{2}kx_m^2sin^2(\omega t+\phi)$$

$$E = K+U$$

$$E = \frac{1}{2}kx_m^2sin^2(\omega t+\phi)+\frac{1}{2}kx_m^2cos^2(\omega t+\phi)$$

$$= \frac{1}{2}kx_m^2(sin^2(\omega t+\phi)+cos^2(\omega t+\phi))$$

* because $sin^2x+cos^2x = 1$
so $E = U+K = \frac{1}{2}kx_m^2$
## Graph