Chapter 4, Problem 4.39P

(a)

To determine

Show that $\tau ={\tau }_0\frac{1}{\pi }{\displaystyle {\int }_0^{\Phi }\frac{d\varphi }{\sqrt{{\sin }^2\left(\Phi /2\right)-{\sin }^2\left(\varphi /2\right)}}}={\tau }_0\frac{2}{\pi }{\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}\sqrt{1-A^2{u}^2}}}$.

Answer to Problem 4.39P

It is proved that $\tau ={\tau }_0\frac{1}{\pi }{\displaystyle {\int }_0^{\Phi }\frac{d\varphi }{\sqrt{{\sin }^2\left(\Phi /2\right)-{\sin }^2\left(\varphi /2\right)}}}={\tau }_0\frac{2}{\pi }{\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}\sqrt{1-A^2{u}^2}}}$.

Explanation of Solution

Write the expression for the potential energy

    $U\left(\varphi \right)=mgl\left(1-\cos \varphi \right)$        (I)

Here, $U$ is the potential energy, $m$ is the mass, $g$ is the acceleration due to gravity, $l$ is the length and $\varphi$ is the angle.

Write the expression for the kinetic energy of the perpendicular

  $K=\frac{1}{2}m{v}^2$        (II)

Here, $K$ is the kinetic energy, $v$ is the linear velocity.

Write the formula for angular velocity

    $\omega =\frac{v}{R}$        (III)

Here, $\omega$ is the angular velocity and $R$ is the radius of the circle.

Where, $\omega =\dot{\varphi }$ and the length of the perpendicular is $R=l$. Equation (II) becomes,

$K=\frac{1}{2}m{l}^2{\dot{\varphi }}^2$        (IV)

The total energy of the system is $E=K+U$.

Substitute equation (I) and (IV) in the above relation

$\begin{array}{c}E=\frac{1}{2}m{l}^2{\dot{\varphi }}^2+mgl\left(1-\cos \varphi \right)\\ \frac{1}{2}m{l}^2{\dot{\varphi }}^2=E-mgl\left(1-\cos \varphi \right)\\ {\dot{\varphi }}^2=\frac{2E}{m{l}^2}-\frac{2g}{l}\left(1-\cos \varphi \right)\\ \dot{\varphi }=\sqrt{\frac{2E}{m{l}^2}-\frac{2g}{l}\left(1-\cos \varphi \right)}\end{array}$

But $t={\displaystyle \int \frac{d\varphi }{\dot{\varphi }}}$

When $\varphi$ becomes maximum, $\dot{\varphi }=0$. And at $\dot{\varphi }=0$ when $E=U\left(\varphi \right)$. But $\dot{\varphi }=0$ when $\varphi =\Phi$.

Therefore, at $\varphi =\Phi$

$E=mgl\left(1-\cos \Phi \right)$

Using the above equation, $\dot{\varphi }$ becomes

$\begin{array}{c}\dot{\varphi }=\sqrt{\frac{2mgl}{m{l}^2}\left(1-\cos \Phi \right)-\frac{2g}{l}\left(1-\cos \varphi \right)}\\ =\sqrt{\frac{2g}{l}\left(1-\cos \Phi \right)-\frac{2g}{l}\left(1-\cos \varphi \right)}\\ =\sqrt{\frac{g}{l}\left[2\left(1-\cos \Phi \right)-2\left(1-\cos \varphi \right)\right]}\\ =\sqrt{\frac{g}{l}\left[2\left(2{\sin }^2\frac{\Phi }{2}\right)-2\left(2{\sin }^2\frac{\varphi }{2}\right)\right]}\end{array}$

$\dot{\varphi }=2\sqrt{\frac{g}{l}\left[{\sin }^2\frac{\Phi }{2}-{\sin }^2\frac{\varphi }{2}\right]}$

Substitute the above equation in $t={\displaystyle \int \frac{d\varphi }{\dot{\varphi }}}$

$\begin{array}{c}t={\displaystyle \int \frac{d\varphi }{2\sqrt{\frac{g}{l}\left[{\sin }^2\frac{\Phi }{2}-{\sin }^2\frac{\varphi }{2}\right]}}}\\ =\frac{1}{2}\sqrt{\frac{l}{g}}{\displaystyle \int \frac{d\varphi }{\sqrt{{\sin }^2\frac{\Phi }{2}-{\sin }^2\frac{\varphi }{2}}}}\end{array}$

Multiply and divide the above equation with $2\pi$ [Note: ${\tau }_0=2\pi \sqrt{\frac{l}{g}}$ ]

$\begin{array}{c}=\frac{2\pi }{2\times 2\pi }\sqrt{\frac{l}{g}}{\displaystyle \int \frac{d\varphi }{\sqrt{{\sin }^2\frac{\Phi }{2}-{\sin }^2\frac{\varphi }{2}}}}\\ t=\frac{1}{4\pi }{\tau }_0\sqrt{\frac{l}{g}}{\displaystyle \int \frac{d\varphi }{\sqrt{{\sin }^2\frac{\Phi }{2}-{\sin }^2\frac{\varphi }{2}}}}\end{array}$

Since $\tau =4t$,

$\begin{array}{l}\tau =4\left[\frac{1}{4\pi }{\tau }_0\sqrt{\frac{l}{g}}{\displaystyle \int \frac{d\varphi }{\sqrt{{\sin }^2\frac{\Phi }{2}-{\sin }^2\frac{\varphi }{2}}}}\right]\\ \tau ={\tau }_0\frac{1}{\pi }{\displaystyle \int \frac{d\varphi }{\sqrt{{\sin }^2\frac{\Phi }{2}-{\sin }^2\frac{\varphi }{2}}}}\end{array}$        (V)

Let $\sin \left(\frac{\varphi }{2}\right)=Au$ and $\sin \left(\frac{\Phi }{2}\right)=A$.

Then,

 $\begin{array}{c}\frac{1}{2}\cos \left(\frac{\varphi }{2}\right)d\varphi =Adu\\ d\varphi =\frac{2Adu}{\cos \left(\frac{\varphi }{2}\right)}\\ d\varphi =\frac{2Adu}{\sqrt{1-{\sin }^2\left(\frac{\varphi }{2}\right)}}\\ d\varphi =\frac{2Adu}{\sqrt{1-A^2{u}^2}}\end{array}$

And,

$\begin{array}{c}{\sin }^2\left(\frac{\Phi }{2}\right)-{\sin }^2\left(\frac{\varphi }{2}\right)=A^2-{\left(Au\right)}^2\\ =A^2\left(1-u^2\right)\end{array}$

When $\varphi =0$, $u=\sin 0=0$

When $\varphi =\Phi$, $\frac{\varphi }{2}=\frac{\Phi }{2}$

$\begin{array}{c}\sin \left(\frac{\varphi }{2}\right)=\sin \left(\frac{\Phi }{2}\right)\\ Au=A\\ u=1\end{array}$

Substituting the above relations in equation (V)

$\begin{array}{c}\tau ={\tau }_0\frac{1}{\pi }{\displaystyle {\int }_0^1\frac{2Adu}{\sqrt{1-A^2{u}^2}}\frac{1}{A\sqrt{1-u^2}}}\\ ={\tau }_0\frac{2}{\pi }{\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}\sqrt{1-A^2{u}^2}}}\end{array}$

Conclusion:

Hence it is proved that $\tau ={\tau }_0\frac{1}{\pi }{\displaystyle {\int }_0^{\Phi }\frac{d\varphi }{\sqrt{{\sin }^2\left(\Phi /2\right)-{\sin }^2\left(\varphi /2\right)}}}={\tau }_0\frac{2}{\pi }{\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}\sqrt{1-A^2{u}^2}}}$.

(b)

To determine

Show that for very small-amplitude period, the equation from part (a) gives familiar result $\tau ={\tau }_0=2\pi \sqrt{\frac{l}{g}}$.

Answer to Problem 4.39P

The assertion is proved that $\tau =2\pi \sqrt{\frac{l}{g}}$ for very small-amplitude period.

Explanation of Solution

Write the equation from part (a),

$\tau ={\tau }_0\frac{1}{\pi }{\displaystyle {\int }_0^{\Phi }\frac{d\varphi }{\sqrt{{\sin }^2\left(\Phi /2\right)-{\sin }^2\left(\varphi /2\right)}}}={\tau }_0\frac{2}{\pi }{\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}\sqrt{1-A^2{u}^2}}}$

When the amplitude, $\Phi$ is very small. Then $A$ is also very small. Therefore, $A^2{u}^2$ can be neglected as it is very small compared with $1$.

The equation becomes,

$\begin{array}{c}\tau ={\tau }_0\frac{2}{\pi }{\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}}}\\ ={\tau }_0\frac{2}{\pi }{\left[{\sin }^{-1}\left(u\right)\right]}_0^1\\ ={\tau }_0\frac{2}{\pi }\left[{\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(0\right)\right]\\ ={\tau }_0\frac{2}{\pi }\left[\frac{\pi }{2}-0\right]\end{array}$

$\begin{array}{c}={\tau }_0\\ \tau =2\pi \sqrt{\frac{l}{g}}\end{array}$

Conclusion:

Hence, the assertion is proved that $\tau =2\pi \sqrt{\frac{l}{g}}$ for very small-amplitude period.

(c)

To determine

Show that $\tau ={\tau }_0\left[1+\frac{1}{4}{\sin }^2\left(\frac{\Phi }{2}\right)\right]$ for small amplitude.

Answer to Problem 4.39P

The assertion is true that for amplitude that are small but not very small $\tau ={\tau }_0\left(1+\frac{1}{4}{\sin }^2\left(\frac{\Phi }{2}\right)\right)$.

Explanation of Solution

Given that $\frac{1}{\sqrt{1-A^2{u}^2}}$ can be approximated to $1+\frac{1}{2}{A}^2{u}^2$ and $A$ is small but not very small.

Write the equation from part (a),

$\tau ={\tau }_0\frac{2}{\pi }{\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}\sqrt{1-A^2{u}^2}}}$

The equation becomes,

$\begin{array}{c}\tau ={\tau }_0\frac{2}{\pi }{\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}}\left(1+\frac{1}{2}{A}^2{u}^2\right)}\\ ={\tau }_0\frac{2}{\pi }\left[{\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}}}+\frac{A^2}{2}{\displaystyle {\int }_0^1\left(\frac{u^2}{\sqrt{1-u^2}}\right)}du\right]\\ ={\tau }_0\frac{2}{\pi }\left[{\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}}}+\frac{A^2}{2}{\displaystyle {\int }_0^1\left(\frac{u^2-1}{\sqrt{1-u^2}}+\frac{1}{\sqrt{1-u^2}}\right)}du\right]\\ ={\tau }_0\frac{2}{\pi }\left[\left(1+\frac{A^2}{2}\right){\displaystyle {\int }_0^1\frac{du}{\sqrt{1-u^2}}}+-\frac{A^2}{2}{\displaystyle {\int }_0^1\sqrt{1-u^2}}du\right]\end{array}$

$\begin{array}{l}={\tau }_0\frac{2}{\pi }\left[\left(1+\frac{A^2}{2}\right)\left(\frac{\pi }{2}\right)+-\frac{A^2}{2}{\left[\frac{u}{2}\sqrt{1-u^2}+\frac{1}{2}{\sin }^{-1}\left(u\right)\right]}_0^1\right]\\ ={\tau }_0\frac{2}{\pi }\left[\left(1+\frac{A^2}{2}\right)\left(\frac{\pi }{2}\right)+-\frac{A^2}{2}\left[0+\frac{1}{2}{\sin }^{-1}\left(1\right)-0-\frac{1}{2}{\sin }^{-1}\left(0\right)\right]\right]\\ ={\tau }_0\frac{2}{\pi }\left[\left(1+\frac{A^2}{2}\right)\left(\frac{\pi }{2}\right)+-\frac{A^2}{2}\left[\frac{1}{2}\frac{\pi }{2}\right]\right]\\ ={\tau }_0\frac{2}{\pi }\left[\left(1+\frac{A^2}{2}-\frac{A^2}{4}\right)\frac{\pi }{2}\right]\end{array}$

$\begin{array}{c}={\tau }_0\left(1+\frac{A^2}{4}\right)\\ \tau ={\tau }_0\left(1+\frac{1}{4}{\sin }^2\left(\frac{\Phi }{2}\right)\right)\end{array}$

Conclusion:

Hence, the assertion is true that for amplitude that are small but not very small $\tau ={\tau }_0\left(1+\frac{1}{4}{\sin }^2\left(\frac{\Phi }{2}\right)\right)$.