Chapter 5.2, Problem 96AYU

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

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

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

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

To determine

To find: The period $T$ (in seconds) of a simple pendulum as a function of its length $l$ (in feet) is given by $T\left(l\right)=2\pi \sqrt{\frac{l}{32.2}}$.

a. Express the length $l$as a function of the period $T$.

Answer to Problem 96AYU

Solution:

a. The length $l$ as a function of the period $T$ is $l=32.2\frac{T^2}{4{\pi }^2}=\frac{8.05T^2}{{\pi }^2}$.

Explanation of Solution

Given:

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

Calculation:

a. Express the length $l$ as a function of the period $T$.

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

$T=2\pi \sqrt{\frac{l}{32.2}}$

$\frac{T}{2\pi }=\sqrt{\frac{l}{32.2}}$

Squaring on both sides, we get

${(\frac{T}{2\pi })}^2=\frac{l}{32.2}$

$\frac{l}{32.2}={(\frac{T}{2\pi })}^2$

$l=32.2{(\frac{T}{2\pi })}^2$

$l=32.2\frac{T^2}{4{\pi }^2}=\frac{8.05T^2}{{\pi }^2}$

To determine

To find: The period $T$ (in seconds) of a simple pendulum as a function of its length $l$(in feet) is given by $T\left(l\right)=2\pi \sqrt{\frac{l}{32.2}}$.

b. How long is a pendulum whose period is 3 seconds?

Answer to Problem 96AYU

Solution:

b. A pendulum whose period 3 seconds is $l=7.3\text{ feets}$.

Explanation of Solution

Given:

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

Calculation:

b. How long is a pendulum whose period is 3 seconds?

We know that $l=32.2\frac{T^2}{4{\pi }^2}=\frac{8.05T^2}{{\pi }^2}$.

We find the length by the given formula.

Given $T=3$ seconds. $l=32.2\frac{T^2}{4{\pi }^2}=\frac{8.05{\left(3\right)}^2}{{\pi }^2}$ $l=\frac{72.45}{9.87}$ $l=7.340426$ $l=7.3 \text{feets}$.