Problem 46 - 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 acceleration due to gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) and the period equals \(1 \mathrm{~s}\), find the length to the nearest \(\mathrm{cm}(100 \mathrm{~cm}=1 \mathrm{~m})\).

Step-by-Step Solution

Answer

The length of the pendulum is 25 cm.

Step 1: Identify Given Variables

We are given the period of the pendulum, \(T = 1\) s, and the acceleration due to gravity, \(g = 9.8\, \text{m/s}^2\). We need to find the length \(L\) of the pendulum.

Step 2: Substitute Known Values into Formula

Substitute the given values into the period formula of the pendulum: \(T = 2 \pi \sqrt{\frac{L}{g}}\). Using \(T = 1\) s and \(g = 9.8\, \text{m/s}^2\), we get \(1 = 2 \pi \sqrt{\frac{L}{9.8}}\).

Step 3: Isolate the Square Root

To isolate the square root, divide both sides by \(2\pi\): \(\frac{1}{2\pi} = \sqrt{\frac{L}{9.8}}\).

Step 4: Square Both Sides

Square both sides of the equation to eliminate the square root: \(\left(\frac{1}{2\pi}\right)^2 = \frac{L}{9.8}\). After simplification, this becomes \(\frac{1}{4\pi^2} = \frac{L}{9.8}\).

Step 5: Solve for L

Multiply both sides by \(9.8\) to solve for \(L\): \(L = 9.8 \times \frac{1}{4\pi^2}\).

Step 6: Calculate L

Calculate \(L\) using the expression: \(L = \frac{9.8}{4\pi^2} \approx \frac{9.8}{39.4784} \approx 0.2482\, \text{m}\).

Step 7: Convert Meters to Centimeters

Convert the length from meters to centimeters: \(0.2482\, \text{m} = 24.82\, \text{cm}\).

Step 8: Round to the Nearest Centimeter

Round \(24.82\, \text{cm}\) to the nearest centimeter, which gives us \(25\, \text{cm}\).

Key Concepts

Pendulum Length

The pendulum is a classic example often used to explore periodic motion in physics. One crucial element of a pendulum is its length, denoted as \( L \). The length of the pendulum directly affects the period \( T \), or the time it takes for the pendulum to complete one full back-and-forth swing.

In the formula, \( T = 2 \pi \sqrt{\frac{L}{g}} \), you'll notice \( L \) is under a square root, demonstrating an important relationship: the period is not linearly proportional to the length. Instead, it increases with the square root of the pendulum's length.

Therefore, doubling the length of the pendulum won't double its period; it will only increase it by a factor of \( \sqrt{2} \). This relationship allows us to calculate the pendulum length when we know the period and the acceleration due to gravity. Understanding this element is essential for calculations in both physics and engineering applications.

Acceleration due to Gravity

The acceleration due to gravity, denoted as \( g \), is a crucial factor in the pendulum period formula. On Earth, \( g \) is approximately \( 9.8 \text{ m/s}^2 \), although it can vary slightly depending on your location due to differences in Earth's density and topography.

This constant is significant because it affects how fast objects accelerate towards the center of the Earth when dropped and also impacts oscillating systems like a pendulum. In the pendulum period formula \( T = 2 \pi \sqrt{\frac{L}{g}} \), \( g \) acts as a divisor inside the square root.

• A higher value of \( g \) results in a shorter period \( T \).
• A lower value of \( g \) results in a longer period \( T \), meaning it takes longer for the pendulum to complete one cycle.

Thus, the pendulum period is inversely related to the square root of the gravitational acceleration. This principle helps explain why pendulums swing differently on planets with stronger or weaker gravity compared to Earth.

Unit Conversion

Unit conversion plays an important role in solving problems involving physical quantities, such as when finding the pendulum's length in the exercise, where it's critical to convert between meters and centimeters. Different scientific fields and geographic regions might rely on various measurement units, making it necessary to understand how to switch between them accurately.

In our formula calculation for the pendulum, once we determine the length \( L \) in meters, converting it to centimeters helps us express it in a more practical unit for real-world applications.

• Remember: there are 100 centimeters in a meter.
• Thus, to convert meters to centimeters, multiply the length in meters by 100.

This conversion is essential for delivering answers in the units most relevant for the problem at hand or required by the exercise.

Square Root Isolation

Isolating the square root is a crucial algebraic technique used in solving the pendulum problem. When given the pendulum formula \( T = 2 \pi \sqrt{\frac{L}{g}} \), we need to rearrange this formula to solve for \( L \).

Here's how the process works:

• Start by dividing both sides of the equation by \( 2\pi \) to isolate the square root: \( \frac{T}{2\pi} = \sqrt{\frac{L}{g}} \).
• Next, square both sides to remove the square root, yielding \( \left(\frac{T}{2\pi}\right)^2 = \frac{L}{g} \).

By carrying out these steps, we transform the equation so that we can solve for \( L \) directly. This step is essential when the desired variable is initially under a square root in the context of physics formulas. Once isolated, you can handle the square root as any other number in further calculations. Mastery of this technique is a gateway to solving many equations beyond pendulum-related problems.