Problem 47 - College Algebra with Corequisite Support

Question

For the following exercises, use the model for the period of a pendulum, \(T\), such that \(T=2 \pi \sqrt{\frac{L}{g}}\), where the length of the pendulum is \(L\) and the acceleration due to gravity is g. If the gravity is \(32 \mathrm{ft} / \mathrm{s}^{2}\) and the period equals \(1 \mathrm{~s}\), find the length to the nearest in. (12 in. = \(1 \mathrm{ft}\) ). Round your answer to the nearest in.

Step-by-Step Solution

Answer

The pendulum length is approximately 3 inches.

Step 1: Understanding the Formula

The period of a pendulum is given by the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.

Step 2: Setting Up the Equation

We need to find the value of \( L \) given that \( T = 1 \) second and \( g = 32 \) ft/s². Substitute these values into the formula: \( 1 = 2\pi \sqrt{\frac{L}{32}} \).

Step 3: Isolating the Square Root

Divide both sides by \( 2\pi \) to isolate the square root term: \( \frac{1}{2\pi} = \sqrt{\frac{L}{32}} \).

Step 4: Squaring Both Sides

Square both sides of the equation to eliminate the square root: \( \left(\frac{1}{2\pi}\right)^2 = \frac{L}{32} \).

Step 5: Solving for L

Multiply both sides by 32 to solve for \( L \): \( L = 32 \cdot \left(\frac{1}{2\pi}\right)^2 \).

Step 6: Calculating L

Calculate: \( L = 32 \cdot \frac{1}{4\pi^2} \approx 32 \cdot 0.025\approx 0.2535 \) ft.

Step 7: Converting to Inches

Convert the length from feet to inches: \( 0.2535 \text{ ft} \times 12 \text{ in/ft} = 3.042 \text{ in} \).

Step 8: Rounding to Nearest Inch

Round the result to the nearest inch: \( 3.042 \text{ in} \) rounds to 3 inches.

Key Concepts

Period of Pendulum

The period of a pendulum, denoted as \(T\), is a fascinating characteristic that measures the time it takes for the pendulum to swing back and forth once. It is crucial in understanding pendulum dynamics and is described by the formula \(T=2 \pi \sqrt{\frac{L}{g}}\). Here, the variables are:

• \(T\): Period of the pendulum (in seconds)
• \(L\): Length of the pendulum (in feet or meters)
• \(g\): Acceleration due to gravity (in feet per second squared or meters per second squared)

This relationship highlights how two factors, the length of the pendulum and the gravitational pull, affect the period. A longer pendulum or a weaker gravitational force results in a longer period. Understanding this formula helps determine missing information when given certain values. For example, if you know the period and gravitational acceleration, you can find the pendulum's length as we do in the exercise.

Acceleration Due to Gravity

The acceleration due to gravity, \(g\), is a constant that significantly influences the motion of a pendulum. It varies depending on location, but on Earth, a common value is approximately \(9.8 \text{ m/s}^2\) or \(32 \text{ ft/s}^2\). This acceleration refers to the rate at which an object speeds up as it falls freely towards the Earth's surface.

• Importance in Pendulum Motion: The strength of gravity directly affects how quickly a pendulum swings; stronger gravity results in a faster swing.
• Practical Application: Using the known gravitational acceleration in an area allows us to calculate or predict pendulum behavior under real-world conditions.

In the exercise, the given value of \(32 \text{ ft/s}^2\) was used, typical for calculations performed in feet. When solving for the pendulum length, this constant plays a vital role in our equations.

Length Calculation

The length calculation of a pendulum is straightforward yet essential to understanding its dynamics and motion. Given the formula \(T=2 \pi \sqrt{\frac{L}{g}}\), we can rearrange it to solve for \(L\) when \(T\) and \(g\) are known.

Steps to Calculate Length

1. Substitute the known values of \(T\) (period) and \(g\) (acceleration due to gravity) into the equation.
2. Isolate \(L\) by dividing both sides by \(2\pi\), then squaring the result to eliminate the square root.
3. Finally, multiply by \(g\) to solve for \(L\).

For example, using the period \(T = 1\) second and \(g = 32 \text{ ft/s}^2\), we follow the process:

• Replace \(1 = 2\pi \sqrt{\frac{L}{32}}\).
• Solve \(L = \left(\frac{1}{2\pi}\right)^2 \times 32\).
• Calculate \(L\approx 0.2535 \text{ feet}\), then convert to inches by multiplying by 12, leading to approximately \(3\) inches (after rounding).

This precise computation allows us to determine the pendulum's length, enabling further insights into its mechanical properties.