Question

In the pendulum problem, $\mathit{\theta }\mathbf{=}\mathit{\alpha }\mathit{s}\mathit{i}\mathit{n}\sqrt{\frac{\mathbf{g}}{\mathbf{l}}}\mathit{t}$ is an approximate solution when the amplitude α is small enough for the motion to be considered simple harmonic. Show that the corresponding exact solution when α is not small is$\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{\alpha }}{\mathbf{2}}\mathit{s}\mathit{n}\sqrt{\frac{\mathbf{g}}{\mathbf{l}}}\mathit{t}\mathbf{,}\mathit{k}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\frac{\mathbf{\alpha }}{\mathbf{2}}\mathbf{\right)}$ is the modulus of the elliptic function. Show that this reduces to the simple harmonic motion solution for small amplitude α

Short answer

The statement has been proven.

Step-by-step solution

Given Information

The equation is $\theta =\alpha \sin \sqrt{\frac{g}{l}}t$.

Definition of elliptic form.

The elliptic form of the integral is defined as $\mathit{K}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\frac{\mathbf{\pi }}{\mathbf{2}}}{\mathbf{\int }}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }}}\mathit{d}\mathit{\theta }$.

Find the value of Integral.

The equation is $\theta =\alpha \sin \sqrt{\frac{g}{l}}t$.

The differential equation is mentioned in the pendulum problem.

The differential equation is$\stackrel{.}{{\theta }^2}=2\frac{g}{l}(\cos \theta -\cos \alpha )$.

Substitute values mentioned below in above equation.

$$\begin{gathered} \xi =\frac{\sin \frac{\theta }{2}}{\sin \frac{\alpha }{2}} \\ \stackrel{.}{\xi }=\frac{{\displaystyle \cos }\frac{\theta }{2}{\displaystyle \stackrel{.}{\theta }}}{{\displaystyle 2}\sin \frac{\alpha }{2}} \end{gathered}$$

The integral becomes as follows

$$\begin{gathered} \frac{4{\sin }^2\frac{\alpha }{2}}{1-{\sin }^2\frac{\alpha }{2}{\xi }^2}\stackrel{.}{\xi }=4{\sin }^2\frac{\alpha }{2}\frac{g}{l}(1-{\xi }^2) \\ \frac{d\xi }{\sqrt{1-{\xi }^2}\sqrt{1-k^2{\xi }^2}}=\sqrt{\frac{g}{l}}dt \\ \underset{0}{\overset{\frac{\sin \frac{\theta }{2}}{\sin \frac{\alpha }{2}}}{\int }}\frac{d\xi }{\sqrt{1-{\xi }^2}\sqrt{1-k^2{\xi }^2}}=\sqrt{\frac{g}{l}}t \\ s{n}^{-1}\left(\frac{\sin \frac{\theta }{2}}{\sin \frac{\alpha }{2}}\right)=\sqrt{\frac{g}{l}}t \end{gathered}$$

Hence$\sin \left(\frac{\theta }{2}\right)=\sin \left(\frac{\alpha }{2}\right)sn\left(\sqrt{\frac{g}{l}}t\right)$.

The formula states the equation mentioned below.

$$\begin{gathered} \theta =\alpha sn\left(\sqrt{\frac{g}{l}}t\right) \\ s{n}^{-1}\left(x\right)=\underset{0}{\overset{x}{\int }}\frac{d\xi }{\sqrt{1-{\xi }^2}} \end{gathered}$$

Integrate the equation mentioned above

$$\begin{gathered} si{n}^{-1}\left(x\right)=s{n}^{-1}\left(x\right) \\ \theta =\alpha \sin \left[\sqrt{\frac{g}{l}}t\right] \end{gathered}$$

Hence, the solution is mentioned below.

The statement has been proven.