Chapter 6.5, Problem 37E

a

To determine

To write a function that represents the distance the pendulum swings on its $n$ th swing.

Answer to Problem 37E

The function that represents the distance the pendulum swings on its $n\text{th}$ swing is $a_n=625{\left(\frac{4}{5}\right)}^{n-1}$

Explanation of Solution

Given information:

The distance (in millimetres) travelled by a swinging pendulum decreases after each swing, as shown in the table.

Swing	1	2	3
Distance (in mm)	625	500	400

  

Calculation:

The distance is given in the geometric sequences: $625,500,400,\mathrm{...}$

  $\text{First term, }{a}_1=625\text{ mm}$

  $\text{Common ratio, }r=\frac{500}{625}=\frac{4}{5}$

The function that represents the distance the pendulum swings on its $n$ th swing is:

  $a_n=a_1{(r)}^{n-1}$

Therefore,

  $a_n=625{\left(\frac{4}{5}\right)}^{n-1}$

Conclusion:

The function that represents the distance the pendulum swings on its $n$ th swing is $a_n=625{\left(\frac{4}{5}\right)}^{n-1}$

b.

To determine

The number of swings, in which the distance $256\text{ mm}$ can be achieved.

Answer to Problem 37E

The pendulum can swing $5$ times to attained the distance $256\text{ mm}$

Explanation of Solution

Given: The distance (in millimetres) travelled by a swinging pendulum decreases after each swing, as shown in the table.

Swing	1	2	3
Distance (in mm)	625	500	400

  

Calculation:

The $n$ th term of the sequence is $a_n=625{\left(\frac{4}{5}\right)}^{n-1}$ , now $a_n=256$

  $\begin{array}{l}{a}_n=625{\left(\frac{4}{5}\right)}^{n-1}\\ 256=625{\left(\frac{4}{5}\right)}^{n-1}\\ \frac{256}{625}={\left(\frac{4}{5}\right)}^{n-1}\\ \left(\frac{4^4}{5^4}\right)={\left(\frac{4}{5}\right)}^{n-1}\\ {\left(\frac{4}{5}\right)}^4={\left(\frac{4}{5}\right)}^{n-1}\end{array}$

Now,

  $\begin{array}{l}4=n-1\\ n=5\end{array}$

Hence,

The pendulum can swing $5$ times to attained the distance $256\text{ mm}$ .