Q. 96 - Precalculus Enhanced with Graphing Utilities

Question

96. Period of a Pendulum The period T (in seconds) of a simple pendulum as a function of its length l (in feet) is

given by

$T\left(l\right)=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{32.2}}$T

(a) Express the length l as a function of the period T.

(b) How long is a pendulum whose period is $3$ seconds?

Step-by-Step Solution

Answer

Part (a). The expression of length as the function of length l  is:

$\mathrm{l}\left(\mathrm{T}\right)={\frac{32.2\mathrm{T}}{4{\mathrm{\pi }}^2}}^2$

Part (b). The length of the pendulum is $7$ meters.

Part (a) Step 1. Express the length l as a function of the period T .

The given function is $T\left(l\right)=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{32.2}}$.

 Expressing the length l  as a function of the period T  is:

$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{32.2}} \\ {\mathrm{T}}^2=4{\mathrm{\pi }}^2\frac{\mathrm{l}}{32.2} \\ 32.2{\mathrm{T}}^2=4{\mathrm{\pi }}^2\mathrm{I} \\ \mathrm{l}\left(\mathrm{T}\right)={\frac{32.2\mathrm{T}}{4{\mathrm{\pi }}^2}}^2 \end{gathered}$$

Part (b) Step 1. Find the time period

Substituting $T=3$ in the function

$I\left(T\right)=32.2\left(\frac{{\mathrm{T}}^2}{4{\mathrm{\pi }}^2}\right)$, we get:

$$\begin{gathered} I\left(T\right)=32.2\left(\frac{{\mathrm{T}}^2}{4{\mathrm{\pi }}^2}\right) \\ l\left(3\right)=32.2\left(\frac{3^2}{4{\mathrm{\pi }}^2}\right) \\ I\left(3\right)=7 \end{gathered}$$