沿X轴方向投影的速度\(v = \dot{x} = A\omega\cos(\omega t + \varphi)\)
沿X轴方向投影的加速度\(a = \ddot{x} = -A\omega^{2}\sin(\omega t + \varphi)\)
回复力$$F = -k\,x = -kA\sin(\omega t + \varphi) = m\,a = -mA\omega^{2}\sin(\omega t + \varphi)$$解得圆频率 \( \omega = \sqrt{\dfrac{k}{m}} \)
周期$$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{m}{k}}$$