# 等時降線

## 解析

${\displaystyle {\frac {\mathrm {d} ^{2}s}{\mathrm {d} t^{2}}}=-k^{2}s}$

${\displaystyle s=s_{0}\cos {kt}}$

${\displaystyle s_{0}}$ 為最低點與釋放點間的弧長，而在最低點時${\displaystyle s=0}$ ，故下滑所需的時間有

${\displaystyle kT_{0}={\frac {\pi }{2}},\ T_{0}={\frac {\pi }{2k}}}$

${\displaystyle {\frac {\mathrm {d} ^{2}s}{\mathrm {d} t^{2}}}=-g\sin \theta }$

${\displaystyle k^{2}s=g\sin \theta }$

${\displaystyle k^{2}{\frac {\mathrm {d} s}{\mathrm {d} \theta }}=g\cos \theta }$

${\displaystyle \mathrm {d} s={\frac {g}{k^{2}}}\cos \theta \,\mathrm {d} \theta }$

${\displaystyle \mathrm {d} x=\cos \theta \,\mathrm {d} s={\frac {g}{k^{2}}}\cos ^{2}\theta \,\mathrm {d} \theta ={\frac {g}{2k^{2}}}\left(1+\cos {2\theta }\right)\mathrm {d} \theta }$

${\displaystyle x=\int {\frac {g}{2k^{2}}}\left(1+\cos {2\theta }\right)\mathrm {d} \theta ={\frac {g}{4k^{2}}}\left(2\theta +\sin {2\theta }\right)+x_{0}}$

${\displaystyle \mathrm {d} y=\sin \theta \,\mathrm {d} s={\frac {g}{k^{2}}}\sin \theta \cos \theta \,\mathrm {d} \theta ={\frac {g}{2k^{2}}}\sin {2\theta }\,\mathrm {d} \theta }$

${\displaystyle y=\int {\frac {g}{2k^{2}}}\sin {2\theta }\,\mathrm {d} \theta =-{\frac {g}{4k^{2}}}\cos {2\theta }+y_{0}}$

${\displaystyle x=r\left(\phi +\sin \phi \right)}$

${\displaystyle y=r\left(1-\cos \phi \right)}$

${\displaystyle k={\sqrt {\frac {g}{4r}}}={\frac {1}{2}}{\sqrt {\frac {g}{r}}}}$

${\displaystyle T_{0}={\frac {\pi }{2k}}=\pi {\sqrt {\frac {r}{g}}}}$

## 阿貝爾力學問題

${\displaystyle mgy_{0}={\frac {1}{2}}mv^{2}+mgy}$

${\displaystyle {\frac {\mathrm {d} s}{\mathrm {d} t}}=\pm {\sqrt {2g\left(y_{0}-y\right)}}}$

${\displaystyle \mathrm {d} t=\pm {\frac {\mathrm {d} s}{\sqrt {2g\left(y_{0}-y\right)}}}=\pm {\frac {1}{\sqrt {2g\left(y_{0}-y\right)}}}{\frac {\mathrm {d} s}{\mathrm {d} y}}\mathrm {d} y}$

${\displaystyle \mathrm {d} t=-{\frac {1}{\sqrt {2g\left(y_{0}-y\right)}}}{\frac {\mathrm {d} s}{\mathrm {d} y}}\mathrm {d} y}$

${\displaystyle T\left(y_{0}\right)=\int _{y=y_{0}}^{y=0}\mathrm {d} t={\frac {1}{\sqrt {2g}}}\int _{0}^{y_{0}}{\frac {1}{\sqrt {y_{0}-y}}}{\frac {\mathrm {d} s}{\mathrm {d} y}}\mathrm {d} y}$

${\displaystyle {\mathcal {L}}\left[T\left(y_{0}\right)\right]={\mathcal {L}}\left[{\frac {1}{\sqrt {y}}}\right]{\mathcal {L}}\left[{\frac {\mathrm {d} s}{\mathrm {d} y}}\right]}$

${\displaystyle {\mathcal {L}}\left[{\frac {\mathrm {d} s}{\mathrm {d} y}}\right]={\sqrt {\frac {2g}{\pi }}}z^{\frac {1}{2}}{\mathcal {L}}\left[T\left(y_{0}\right)\right]}$

${\displaystyle {\mathcal {L}}\left[T_{0}\right]=T_{0}{\mathcal {L}}\left[1\right]={\frac {1}{z}}T_{0}}$

${\displaystyle {\mathcal {L}}\left[{\frac {\mathrm {d} s}{\mathrm {d} y}}\right]={\sqrt {\frac {2g}{\pi }}}z^{\frac {1}{2}}{\mathcal {L}}\left[T_{0}\right]={\sqrt {\frac {2g}{\pi }}}T_{0}z^{-{\frac {1}{2}}}}$

${\displaystyle {\frac {\mathrm {d} s}{\mathrm {d} y}}={\frac {\sqrt {2g}}{\pi }}T_{0}y^{-{\frac {1}{2}}}}$

${\displaystyle \mathrm {d} s={\frac {\sqrt {2g}}{\pi }}T_{0}y^{-{\frac {1}{2}}}\mathrm {d} y}$

${\displaystyle \mathrm {d} y=\sin \theta \mathrm {d} s}$ ，易得

${\displaystyle y^{\frac {1}{2}}={\frac {\sqrt {2g}}{\pi }}T_{0}\sin \theta }$

${\displaystyle \mathrm {d} \left(y^{\frac {1}{2}}\right)={\frac {1}{2}}y^{-{\frac {1}{2}}}\mathrm {d} y={\frac {\sqrt {2g}}{\pi }}T_{0}\cos \theta \,\mathrm {d} \theta }$

${\displaystyle \mathrm {d} s={\frac {4g^{2}}{\pi ^{2}}}T_{0}^{2}\cos \theta \,\mathrm {d} \theta }$

${\displaystyle \mathrm {d} s={\frac {g}{k^{2}}}T_{0}^{2}\cos \theta \,\mathrm {d} \theta }$

${\displaystyle T_{0}^{2}={\cfrac {\pi ^{2}}{4k^{2}g}}}$

${\displaystyle T_{0}=\pi {\sqrt {\frac {r}{g}}}}$