Chapter 8.8, Problem 32E

(a)

To determine

To Prove: The expansion of the integrand as the binomial series and then use ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx}=\frac{1:3:\mathrm{5....}\left(2n-1\right)}{2\cdot 4\cdot \mathrm{6.......2}n}\cdot \frac{\pi }{2}$ to shows that $T=2\pi \sqrt{\frac{L}{g}}\left[1+\frac{1^2}{2^2}{k}^2+\frac{1^2\cdot 3^2}{2^2\cdot 4^2}{k}^4+\frac{1^2\cdot 3^2\cdot 5^2}{2^2\cdot 4^2\cdot 6^2}{k}^6+\mathrm{.....}\right]$

Explanation of Solution

Given:

For ${\theta }_0$ not too large the approximation $T=2\pi \sqrt{\frac{L}{g}}$ is obtained by the use of only first term of the series by the use two term $T=2\pi \sqrt{\frac{L}{g}}\left(1+\frac{1}{4}{k}^2\right)$ .

Calculation:

Consider for the binomial series is

  ${\left(1+x\right)}^r={\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}r\\ n\end{array}\right)x^n}=1+rx+\frac{r\left(r-1\right)x^2}{2!}+\frac{r\left(r-1\right)\left(r-2\right)}{3!}{x}^3+\mathrm{..}$

Then,

  $\begin{array}{c}T=4\sqrt{\frac{L}{g}}{\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{1-k^2{\sin }^{2}x}}}\\ =4\sqrt{\frac{L}{g}}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\left(1-k^2{\sin }^{2}x\right)}^{\frac{-1}{2}}dx}\\ =4\sqrt{\frac{L}{g}}{\displaystyle {\int }_0^{\frac{\pi }{2}}\left(\begin{array}{c}\frac{-1}{2}\\ n\end{array}\right){\left(-k^2\right)}^n}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx}\end{array}$

Consider the solution of the part is ${\displaystyle {\int }_0^{\frac{\pi }{2}}\left(\begin{array}{c}\frac{-1}{2}\\ n\end{array}\right){\left(-k^2\right)}^n}$ is,

  ${\displaystyle {\int }_0^{\frac{\pi }{2}}\left(\begin{array}{c}\frac{-1}{2}\\ n\end{array}\right){\left(-k^2\right)}^n}=1+\left(\frac{-1}{2}\right)\left(-k^2\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{1\cdot 2}{\left(-k^2\right)}^2+\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{1\cdot 2\cdot 3}{\left(-k^2\right)}^3+\mathrm{..}$

  ${\displaystyle {\int }_0^{\frac{\pi }{2}}\left(\begin{array}{c}\frac{-1}{2}\\ n\end{array}\right){\left(-k^2\right)}^n}=1+\frac{1}{2}{k}^2+\frac{1\cdot 3}{2\cdot 4}{k}^4+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}{k}^6+\mathrm{....}$

  ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx}{\displaystyle {\int }_0^{\frac{\pi }{2}}\left(\begin{array}{c}\frac{-1}{2}\\ n\end{array}\right){\left(-k^2\right)}^n}={\displaystyle \sum _{n=0}^{\infty }\left(\frac{1\cdot 3\cdot 5\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(2n-1\right)}{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot 2n}{k}^{2n}\right)}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx}$

  ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx}{\displaystyle {\int }_0^{\frac{\pi }{2}}\left(\begin{array}{c}\frac{-1}{2}\\ n\end{array}\right){\left(-k^2\right)}^n}={\displaystyle \sum _{n=0}^{\infty }\left(\frac{1^2\cdot 3^2\cdot 5^2\cdot \cdot \cdot {\left(2n-1\right)}^2}{2^2\cdot 4^2\cdot 6^2\cdot \cdot \cdot \cdot 2n^2}{k}^{2n}\right)}\cdot k^{2n}\cdot \frac{\pi }{2}$

Solve further as,

  $\begin{array}{l}4\sqrt{\frac{L}{g}}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx}{\displaystyle {\int }_0^{\frac{\pi }{2}}\left(\begin{array}{c}\frac{-1}{2}\\ n\end{array}\right){\left(-k^2\right)}^n}=4\sqrt{\frac{L}{g}}{\displaystyle \sum _{n=0}^{\infty }\left(\frac{1^2\cdot 3^2\cdot 5^2\cdot \cdot \cdot {\left(2n-1\right)}^2}{2^2\cdot 4^2\cdot 6^2\cdot \cdot \cdot \cdot 2n^2}{k}^{2n}\right)}\cdot k^{2n}\cdot \frac{\pi }{2}\\ T=2\pi \sqrt{\frac{L}{g}}\cdot \left(1+\frac{1^2}{2^2}{k}^2+\frac{1^2\cdot 3^2}{2^2\cdot 4^2}{k}^4+\frac{1^2\cdot 3^2\cdot 5^2}{2^2\cdot 4^2\cdot 6^2}\cdot k^6\mathrm{...}\right)\end{array}$

Hence, proved.

(b)

To determine

To Prove: The fact that all the terms in the series after the first one have coefficients that most are at most $\frac{1}{4}$ to prove that the $2\pi \sqrt{\frac{L}{g}}\left(1+\frac{1}{4}{k}^2\right)\le T\le 2\pi \sqrt{\frac{L}{g}}\frac{4-3k^2}{4-4k^2}$ .

Answer to Problem 32E

The required graph is shown in Figure 2

Explanation of Solution

Given:

The thermal coefficient is $\alpha =\text{0}\text{.039}/\text{°C}$ and the resistivity is ${\rho }_{20}=1.7\times {10}^{-8}\Omega \cdot \text{m}$ .

Calculation:

Consider the expression is,

  $T=2\pi \sqrt{\frac{L}{g}}\cdot \left(1+\frac{1^2}{2^2}{k}^2+\frac{1^2\cdot 3^2}{2^2\cdot 4^2}{k}^4+\frac{1^2\cdot 3^2\cdot 5^2}{2^2\cdot 4^2\cdot 6^2}\cdot k^6\mathrm{...}\right)$

Then,

  $T\ge 2\pi \sqrt{\frac{L}{g}}\cdot \left(1+\frac{1}{4}{k}^2\right)$

Consider that the inequality occurs for the case $k=0$ then,

  $2\pi \sqrt{\frac{L}{g}}\cdot \left(1+\frac{1}{4}{k}^2\right)\le T$

Since, the coefficient of the term is less than $\frac{1}{4}$ . Then,

  $\begin{array}{l}2\pi \sqrt{\frac{L}{g}}\cdot \left(1+\frac{1}{4}{k}^2+\frac{1}{4}{k}^4+\frac{1}{4}{k}^6\mathrm{...}\right)\le T\\ T\le 2\pi \sqrt{\frac{L}{g}}\cdot \left(\frac{4-3k^2}{4-4k^2}\right)\end{array}$

Hence, Proved.

(c)

To determine

To Find: The way in which the given condition compare to the estimate $T=2\pi \sqrt{\frac{L}{g}}$ and what if ${\theta }_0=42°$ .

Answer to Problem 32E

The time period the value is close to ${\theta }_0=10$ and $2.07154\le T\le 2.08104$ .

Explanation of Solution

Calculation:

Consider that the $k=\sin \frac{1}{2}{\theta }_0$

For ${\theta }_0=10°$ , then $k=\sin 5°$ is equal to 0.0871.

Then,

  $\begin{array}{l}2\pi \sqrt{\frac{L}{g}}\cdot \left(1+\frac{1}{4}\cdot k^2\right)\le T\le 2\pi \sqrt{\frac{L}{g}}\cdot \left(\frac{4-3k^2}{4-4k^2}\right)\\ 2\pi \sqrt{\frac{L}{g}}\cdot \left(1+\frac{1}{4}\cdot {\left(0.0871\right)}^2\right)\le T\le 2\pi \sqrt{\frac{L}{g}}\cdot \left(\frac{4-3{\left(0.0871\right)}^2}{4-4{\left(0.0871\right)}^2}\right)\\ 2.01090\le T\le 2.01093\end{array}$

For ${\theta }_0=42°$ , then $k=\sin 21°$ is equal to 0.3584.

Then,

  $\begin{array}{l}2\pi \sqrt{\frac{L}{g}}\cdot \left(1+\frac{1}{4}\cdot k^2\right)\le T\le 2\pi \sqrt{\frac{L}{g}}\cdot \left(\frac{4-3k^2}{4-4k^2}\right)\\ 2\pi \sqrt{\frac{L}{g}}\cdot \left(1+\frac{1}{4}\cdot {\left(0.3584\right)}^2\right)\le T\le 2\pi \sqrt{\frac{L}{g}}\cdot \left(\frac{4-3{\left(0.3584\right)}^2}{4-4{\left(0.3584\right)}^2}\right)\\ 2.07154\le T\le 2.08104\end{array}$

The n,

  $\begin{array}{c}T=2\pi \sqrt{\frac{L}{g}}\\ =2\pi \sqrt{\frac{1}{9.8}}\\ =2.00709\end{array}$

For the above value of the time period the value is close to ${\theta }_0=10$ .