Problem 47 - College Algebra

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 \(32 \mathrm{ft} /{ }^{2}\) and the period equals \(1 \mathrm{~s}\), fi d the length to the nearest in. (12 in. = \(1 \mathrm{ft}\) ). Round your answer to the nearest in.

Step-by-Step Solution

Answer

The length of the pendulum is approximately 10 inches.

Step 1: Understand the Problem

We need to find the length of the pendulum, denoted as \(L\), given that the period \(T\) is 1 second and the acceleration due to gravity \(g\) is 32 ft/s². The formula for the period of a pendulum is provided as \(T = 2\pi \sqrt{\frac{L}{g}}\).

Step 2: Substitute Known Values into the Formula

Substitute \(T = 1\) second and \(g = 32\) ft/s² into the period formula: \(1 = 2\pi \sqrt{\frac{L}{32}}\).

Step 3: Solve for \(L\)

Firstly, divide both sides by \(2\pi\): \(\frac{1}{2\pi} = \sqrt{\frac{L}{32}}\). Then, square both sides to eliminate the square root: \(\left(\frac{1}{2\pi}\right)^2 = \frac{L}{32}\). Now, multiply both sides by 32 to solve for \(L\): \(L = 32 \times \left(\frac{1}{2\pi}\right)^2\).

Step 4: Perform Calculations

Calculate \(\left(\frac{1}{2\pi}\right)^2\): 1. Calculate \(2\pi\): approximately 6.2832.2. Compute \(\frac{1}{2\pi}\): approximately 0.1592.3. Square the result: \(\approx 0.0253\).4. Multiply by 32 to find \(L\): \(L \approx 32 \times 0.0253 \approx 0.8096\) feet.

Step 5: Convert Length to Inches

Since there are 12 inches in a foot, convert \(L\) to inches: \(L = 0.8096 \times 12 \approx 9.715\) inches.

Step 6: Round to the Nearest Inch

Round 9.715 inches to the nearest whole number. The length of the pendulum to the nearest inch is 10 inches.

Key Concepts

Acceleration due to Gravity

The acceleration due to gravity is a key factor in understanding pendulum behavior. It signifies how fast an object accelerates towards the Earth. On our planet, gravity accelerates objects at approximately 9.8 meters per second squared (m/s²), or equivalently 32 feet per second squared (ft/s²) as used in the pendulum exercise. This constant is crucial because it influences how quickly the pendulum swings. It ensures that all objects, regardless of their mass, fall at the same rate when only gravity is acting upon them. Knowing the value of gravity in the environment where the pendulum is swinging allows us to accurately calculate the pendulum's period using the given formula.

Unit Conversion

Unit conversion is important for solving problems in physics, especially when different systems of measurement are used. In this exercise, the length was first calculated in feet. Later, it was converted to inches to give the answer in a more commonly used unit. Here are the steps for unit conversion:

• Identify the conversion factor. In this case, 12 inches equals 1 foot.
• Use multiplication or division to convert between units, maintaining consistency in your calculations.

By multiplying the pendulum length in feet by the conversion factor (12 inches per foot), the length was successfully converted to inches, making the value more comprehensible for everyday use.

Square Root

The square root is a mathematical concept symbolized by the radical sign, \( \sqrt{} \). It's used to find a number which, when multiplied by itself, gives the original number. For pendulum calculations, the square root comes into play in the formula \( T = 2 \pi \sqrt{\frac{L}{g}} \), where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. To solve such equations, you need to isolate the square root term and then square both sides to simplify the equation. In essence, solving involves removing the square root so you can find the actual length of the pendulum. It's an essential step that helps transition from a theoretical formula to practical usable calculations.

Mathematical Modeling

Mathematical modeling is the process of using mathematical formulas to represent real-world scenarios. This is central to physics, allowing us to predict how systems behave. The pendulum formula, \( T = 2\pi \sqrt{\frac{L}{g}} \), is a fine example of such modeling. It predicts the period based on length and gravity, abstracting the physical phenomenon into a precise equation.
Mathematical models like this help simplify complex systems and provide clear insights into how variables interact. In this problem, by substituting known values into the model, one can solve for the unknown variable \( L \). This approach shows the power of mathematics to model and solve everyday physical problems with elegance and precision.