Chapter 9.5, Problem 50PPSE

Solution

a.

Tofind the swing on which the length of the swing is shorter than 50 cm for the first time.

The swing of the pendulum will be less than 50 com for the first time on 70^th swing.

Given information:

The first swing of the pendulum is of length 100 cm and the return swing is 99 cm.

Calculation:

The swing of the pendulum forms an infinite geometric series with $a_1=100$ and $r=\frac{99}{100}$ .

The general term can be written as:

  $\begin{array}{c}{a}_n=a_1{r}^{n-1}\\ a_n=100{\left(0.99\right)}^{n-1}\end{array}$

Now, to find the swing on which the length of the swing is less than 50 cm for the first time, solve the inequality, $100{\left(0.99\right)}^{n-1}<50$ .

  $\begin{array}{c}100{\left(0.99\right)}^{n-1}<50\\ {\left(0.99\right)}^{n-1}<0.5\\ n-1<{\log }_{0.99}\left(0.5\right)\\ n>69.967\end{array}$

This shows that the swing of the pendulum will be less than 50 com for the first time on 70^th swing.

b.

To find the distance travelled by the pendulum when it comes to rest.

  $10000$ cm.

Given information:

The first swing of the pendulum is of length 100 cm and the return swing is 99 cm.

Calculation:

The swing of the pendulum forms an infinite geometric series with $a_1=100$ and $r=\frac{99}{100}$ .

Substitute the values $a_1=100$ and $r=0.99$ in the formula $S=\frac{a_1}{1-r}$ to find the distance travelled.

  $\begin{array}{c}S=\frac{a_1}{1-r}\\ =\frac{100}{1-0.99}\\ =\frac{100}{0.01}\\ =10000\end{array}$

Therefore, the total distance travelled by the pendulum is $10000$ cm.