Problem 61 - College Algebra

Question

For the following exercises, determine the function described and then use it to answer the question. The period \(T,\) in seconds, of a simple pendulum as a function of its length \(l,\) in feet, is given by \(T(l)=2 \pi \sqrt{\frac{l}{32.2}} .\) Express \(l\) as a function of \(T\) and determine the length of a pendulum with period of 2 seconds.

Step-by-Step Solution

Answer

The pendulum length is approximately 3.26 feet.

Step 1: Understand the Given Function

We are given the function \(T(l) = 2 \pi \sqrt{\frac{l}{32.2}} \), where \(T\) is the period in seconds, and \(l\) is the length in feet.

Step 2: Isolate the Square Root

To express \(l\) as a function of \(T\), we start by isolating the square root in the formula. Divide both sides by \(2\pi\):\[ \frac{T}{2\pi} = \sqrt{\frac{l}{32.2}} \]

Step 3: Square Both Sides

Eliminate the square root by squaring both sides of the equation:\[ \left(\frac{T}{2\pi}\right)^2 = \frac{l}{32.2} \]

Step 4: Solve for l

Multiply both sides by 32.2 to solve for \(l\):\[ l = 32.2 \left(\frac{T}{2\pi}\right)^2 \]

Step 5: Substitute T = 2 seconds and Calculate

Substitute \(T = 2\) into the equation for \(l\):\[ l = 32.2 \left(\frac{2}{2\pi}\right)^2 = 32.2 \left(\frac{1}{\pi}\right)^2 \approx 32.2 \times 0.1013 \approx 3.26 \text{ feet} \]

Key Concepts

Function Inversion

Function inversion involves rewriting a function to express the original input variable of the function in terms of the output variable. In simpler terms, it's reversing the roles of the input and output. This is very useful in scenarios where you have the output and would like to find the input that led to it.

For example, in the given problem, we are initially provided with the function for the pendulum's period based on its length, \( T(l) = 2 \pi \sqrt{\frac{l}{32.2}} \). If we want to find out how long the pendulum needs to be for a desired period, it's necessary to invert this function, turning the equation around to solve for \( l \).

The process starts with isolating the term with the variable \( l \), and manipulatively solving the equation step-by-step, where you divide, square, and then isolate the variable with basic algebraic manipulation. This teaches us how to think in reverse to get the relationship between two related variables.

Problem Solving

Problem solving in mathematics often involves breaking down a complex task into several manageable steps. When faced with an equation to reverse, as seen here, it's essential to clearly understand every part of the given function.

• **Understand and Analyze:** Look at the equation you're given. In this case, it’s important to interpret \( T(l) = 2 \pi \sqrt{\frac{l}{32.2}} \) as a relationship where you know \( T \) and need to derive \( l \).
• **Manipulate Algebraically:** Apply algebraic operations to rearrange the equation to find the desired variable. Dividing, squaring, multiplying, or applying other operations in the logical direction is key.
• **Solution Verification:** Finally, verify the result obtained from your manipulated equation. Substitute back into your original condition to check the correctness and relevancy of the solution.

This approach helps ensure you don’t just follow mechanical steps, but instead you gain a deeper comprehension of the why behind each step. It’s a vital skill across various mathematical problems and real-world tasks.

Pendulum Period Formula

The pendulum period formula is essential in physics, particularly in understanding the behavior of pendulums. This formula, \( T(l) = 2 \pi \sqrt{\frac{l}{32.2}} \), describes how the period \( T \)—the time it takes for one complete cycle of the pendulum—is dependent on the length \( l \) of the pendulum arm.

- This relationship shows that a longer pendulum has a longer period, which means it takes more time to swing back and forth. - Meanwhile, the formula’s constant terms relate to the physics behind acceleration due to gravity, as 32.2 ft/s² is the approximate acceleration due to gravity on Earth.

Setting the formula in place allows us to solve various practical and theoretical problems, like draping pendulums in clocks, or even in scientific experiments where timing precision is critical. When given this function, manipulating it as shown earlier to find different elements underscores its wide utility in technical and education fields alike.