繩嘅振動

${\displaystyle v={\sqrt {\frac {T}{\mu }}}}$

證明

${\displaystyle \tan \alpha ={\frac {T_{1}\sin \alpha }{T_{1}\cos \alpha }}=\left.{\frac {\partial y(x,t)}{\partial x}}\right|_{(x,t)}}$

${\displaystyle \tan \beta ={\frac {T_{2}\sin \beta }{T_{2}\cos \beta }}=\left.{\frac {\partial y(x,t)}{\partial x}}\right|_{(x+dx,t)}}$

${\displaystyle T_{1}\cos \alpha =T_{2}\cos \beta =T}$

${\displaystyle T_{1}\sin \alpha =T\left.{\frac {\partial y(x,t)}{\partial x}}\right|_{(x,t)}}$

${\displaystyle T_{2}\sin \beta =T\left.{\frac {\partial y(x,t)}{\partial x}}\right|_{(x+dx,t)}}$

${\displaystyle F_{net}=T_{2}\sin \beta -T_{1}\sin \alpha =T(\left.{\frac {\partial y(x,t)}{\partial x}}\right|_{(x+dx,t)}-\left.{\frac {\partial y(x,t)}{\partial x}}\right|_{(x,t)})=T(y_{x}(x+dx,t)-y_{x}(x,t))}$

${\displaystyle F_{net}=ma=\mu dx{\frac {\partial ^{2}y}{\partial t^{2}}}}$

${\displaystyle T(y_{x}(x+dx,t)-y_{x}(x,t))=\mu dx{\frac {\partial ^{2}y}{\partial t^{2}}}}$

${\displaystyle {\frac {y_{x}(x+dx,t)-y_{x}(x,t)}{dx}}={\frac {\mu }{T}}{\frac {\partial ^{2}y}{\partial t^{2}}}}$

${\displaystyle {\frac {\partial ^{2}y}{\partial x^{2}}}={\frac {\mu }{T}}{\frac {\partial ^{2}y}{\partial t^{2}}}}$

${\displaystyle {\frac {1}{v^{2}}}={\frac {\mu }{T}}}$

${\displaystyle y=A\sin(kx-\omega t+\phi )}$

${\displaystyle -k^{2}A\sin(kx-\omega t+\phi )=-{\frac {\mu }{T}}\omega ^{2}A\sin(kx-\omega t+\phi )}$

${\displaystyle {\frac {\omega ^{2}}{k^{2}}}={\frac {T}{\mu }}}$

${\displaystyle v^{2}={\frac {T}{\mu }}}$