高观点下的高中物理：修订间差异

对于简谐振动，我们有动力学方程<math>ma = -kx</math>

即<math>m\ddot{x} + kx = 0</math>

不妨设<math>x = Ae^{\textup{i}\omega t}</math>

则<math>-mA\omega^2e^{\textup{i}\omega t} + kAe^{\textup{i}\omega t} = 0</math>

解得<math>\omega = \pm\sqrt{\frac{k}{m}}</math>

所以<math>x(t) = C_1e^{\textup{i}\sqrt{\frac{k}{m}}t} + C_2e^{-\textup{i}\sqrt{\frac{k}{m}}t}</math>

<math>a(t) = -\frac{k}{m}C_1e^{\textup{i}\sqrt{\frac{k}{m}}t} -\frac{k}{m}C_2e^{-\textup{i}\sqrt{\frac{k}{m}}t}</math>

可带入特解<math>x(0) = A,a(0) = -\frac{kx(0)}{m}</math>

解得<math>C_1 = C_2 = \frac{1}{2}A</math>

所以<math>x(t) = \frac{1}{2}A\left(e^{\textup{i}\sqrt{\frac{k}{m}}t} + e^{-\textup{i}\sqrt{\frac{k}{m}}t}\right)</math>

通过欧拉公式化简可以得到<math>x(t) = A\cos(\omega t),\omega = \sqrt{\frac{k}{m}}</math>

所以<math>T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}</math>