Problem 58 - Algebra for College Students

Question

Solve the formula \(T=2 \pi \sqrt{\frac{L}{32}}\) for \(L\). (Remember that in this formula, which was used in Section \(5.2, T\) represents the period of a pendulum expressed in seconds, and \(L\) represents the length of the pendulum in feet.)

Step-by-Step Solution

Answer

\(L = 32 \left(\frac{T}{2\pi}\right)^2\).

Step 1: Understanding the Formula

We start with the formula for the period of a pendulum: \( T = 2 \pi \sqrt{\frac{L}{32}} \). Our goal is to solve this formula for \(L\), which represents the length of the pendulum.

Step 2: Isolating the Square Root

Divide both sides of the equation by \(2\pi\) to isolate the square root: \( \frac{T}{2\pi} = \sqrt{\frac{L}{32}} \). This steps simplifies the equation to make the square root alone on one side.

Step 3: Removing the Square Root

To remove the square root, square both sides of the equation: \( \left(\frac{T}{2\pi}\right)^2 = \frac{L}{32} \). This step allows us to express the relationship without the square root, simplifying the next step.

Step 4: Solving for L

Multiply both sides by 32 to solve for \(L\): \( L = 32 \left(\frac{T}{2\pi}\right)^2 \). This gives us the expression for \(L\) as a function of \(T\).

Key Concepts

Pendulum Period Formula

The pendulum period formula provides a way to calculate the time it takes for a pendulum of a certain length to complete one full swing. This is represented by the formula:\[ T = 2 \, \pi \, \sqrt{\frac{L}{32}} \]In this equation:

• \(T\) is the period, the time of one complete oscillation, measured in seconds.
• \(L\) is the length of the pendulum, given in feet.

This relationship comes from the theory of simple harmonic motion. The pendulum's period depends solely on its length, assuming the pendulum swings with a small amplitude in a uniform gravitational field. By solving the formula for different variables, you can explore how changes in length affect the pendulum's behavior.

Square Root Isolation

Square root isolation refers to the process of separating the square root term on one side of the equation to solve for an unknown variable. Let's look at our problem:To solve for \(L\), we first isolate the square root by dividing both sides by \(2\pi\):\[ \frac{T}{2\pi} = \sqrt{\frac{L}{32}} \]This step keeps the equation balanced while allowing the square root to stand alone. Isolating the square root is crucial because it sets up for the next step, which involves removing the square root by squaring both sides of the equation. This method is frequently used in mathematics to simplify equations and solve for unknowns efficiently. By squaring both sides, we can work with quantities directly and eliminate the root.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to solve for a particular variable. Once the square root is isolated, we can perform algebraic operations on both sides to find \(L\) in terms of \(T\).After isolating the square root and squaring both sides, we have:\[ \left(\frac{T}{2\pi}\right)^2 = \frac{L}{32} \]The next algebraic step is to multiply both sides by 32 to solve for \(L\):\[ L = 32 \left(\frac{T}{2\pi}\right)^2 \]By doing this, we have isolated \(L\)—our variable of interest—and expressed it as a function of \(T\). Each manipulation step is designed to maintain the balance of the equation. Through these transformations, we solve effectively for the length \(L\), demonstrating how mathematical principles can be applied to uphold relationships between physical quantities.