Chapter 4.7, Problem 39E

(a)

To determine

The period of a 1-meter pendulum on Earth if the acceleration due to gravity at earth’s surface is 9.8 meters per second squared?

Answer to Problem 39E

  $T\simeq 2.01\text{ sec}$

Explanation of Solution

Given information: Physics: The period of a pendulum (the time required for one back and forth swing) can be determined by the formula $T=2\pi \sqrt{\frac{l}{g}}$ . In this formula, T represents the period, l represents the length of the pendulum and g represents acceleration due to gravity.

Calculation:

  $T=2\pi \sqrt{\frac{l}{g}}$

  $l=1,g=9.8$

  $\begin{array}{l}T=2\pi \sqrt{\frac{1}{9.8}}\\ T\simeq 2.01\text{ sec}\end{array}$

(b)

To determine

To calculate: The period of the pendulum on Venus, suppose the acceleration due to gravity on the surface of Venus is 8.9 meters per second squared.

Answer to Problem 39E

  $T\simeq 2.11\text{ sec}$

Explanation of Solution

Given information: Physics: The period of a pendulum (the time required for one back and forth swing) can be determined by the formula $T=2\pi \sqrt{\frac{l}{g}}$ . In this formula, T represents the period, l represents the length of the pendulum and g represents acceleration due to gravity.

Calculation:

  $T=2\pi \sqrt{\frac{l}{g}}$

  $l=1,g=8.9$

  $\begin{array}{l}T=2\pi \sqrt{\frac{1}{8.9}}\\ T\simeq 2.11\text{ sec}\end{array}$

(c)

To determine

How must the length of the pendulum be changed to double the period?

Answer to Problem 39E

The length is multiplied by 4 to double the period

Explanation of Solution

Given information: Physics: The period of a pendulum (the time required for one back and forth swing) can be determined by the formula $T=2\pi \sqrt{\frac{l}{g}}$ . In this formula, T represents the period, l represents the length of the pendulum and g represents acceleration due to gravity.

Calculation:

  $T=2\pi \sqrt{\frac{l}{g}}$

  $n\text{ is}$ the new length

  $\begin{array}{l}2\left(2\pi \sqrt{\frac{l}{g}}\right)=2\pi \sqrt{\frac{n}{g}}\\ 4\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{n}{g}}\\ 2\sqrt{\frac{l}{g}}=\sqrt{\frac{n}{g}}\\ 4\frac{l}{g}=\frac{n}{g}\\ 4l=n\end{array}$

This means the length is multiplied by 4 to double the period.