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

Step-by-Step Solution

Answer

The length of the pendulum is approximately 25 cm.

Step 1: Identify given values

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

Step 2: Rearrange the formula for L

The model for the period of a pendulum is given by:\[ T = 2\pi \sqrt{\frac{L}{g}} \]We need to solve for \( L \). Start by dividing both sides by \( 2\pi \):\[ \frac{T}{2\pi} = \sqrt{\frac{L}{g}} \]

Step 3: Eliminate the square root

Square both sides of the equation to get rid of the square root:\[ \left(\frac{T}{2\pi}\right)^2 = \frac{L}{g} \]

Step 4: Solve for L

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

Step 5: Substitute known values

Substitute \( T = 1 \text{ s} \) and \( g = 9.8 \text{ m/s}^2 \) into the equation:\[ L = 9.8 \left(\frac{1}{2\pi}\right)^2 \]

Step 6: Calculate L and convert to cm

Calculate the value of \( L \) using the equation:\[ L = 9.8 \left(\frac{1}{6.2832}\right)^2 \approx 9.8 \times 0.0253 \approx 0.248 \text{ m} \]Convert \( L \) to centimeters:\[ 0.248 \text{ m} \times 100 = 24.8 \text{ cm} \]Rounding to the nearest centimeter, \( L \approx 25 \text{ cm} \).

Key Concepts

Pendulum Length Calculation

To calculate the length of a pendulum, we use the formula for the period of a pendulum, given by: \[ T = 2 \pi \sqrt{\frac{L}{g}} \]In this context, the period \( T \) is the time it takes for one complete cycle or oscillation. The task is to solve this formula for \( L \), the length of the pendulum.

• Initial Set-up: Given the values, we have the period \( T = 1 \text{ s} \) and gravity \( g = 9.8 \text{ m/s}^2 \).
• Formula Rearrangement: To rearrange the formula, divide both sides by \( 2\pi \), leading to \( \frac{T}{2\pi} = \sqrt{\frac{L}{g}} \).
• Solving for \( L \): Square both sides of the equation, \( \left(\frac{T}{2\pi}\right)^2 = \frac{L}{g} \), and multiply by \( g \) to isolate \( L \) giving \( L = g \left(\frac{T}{2\pi}\right)^2 \).

With these steps, we can substitute our known values into the equation to find the pendulum's length.

Gravity in Pendulum Motion

Gravity plays a key role in the movement of a pendulum. The force of gravity is what makes the pendulum swing back and forth in regular periodic motion. In our formula, gravity is represented by \( g \).

• Value of \( g \): In many physics problems, the acceleration due to gravity is approximated as \( 9.8 \text{ m/s}^2 \).
• Impact on Motion: Gravity constantly accelerates the pendulum towards its resting position, influencing the speed and period of the swing.

Using the gravitational constant ensures our calculations reflect realistic pendulum behavior on Earth. The stronger the gravity, the shorter the period of the pendulum's swing if the length is kept constant, providing vital information for accurately determining the pendulum's characteristics.

Algebraic Manipulation

Algebraic manipulation is essential in solving equations, especially when isolating variables. Let's look at how this technique helps us solve for the pendulum length \( L \):

• Rearranging Formulas: The starting equation, \( T = 2\pi\sqrt{\frac{L}{g}} \), requires first isolating the radical expression by dividing both sides by \( 2\pi \).
• Removing a Square Root: To eliminate the square root, we square each side of the equation, crucial in ensuring that the equation remains balanced.
• Final Steps: To fully solve for \( L \), multiply by \( g \) on both sides. This manipulation reveals \( L = g \left(\frac{T}{2\pi}\right)^2 \).

Every step uses algebraic manipulation to maintain equality. Mastering these techniques is key to tackling a wide array of problems not only in pendulum motion but in many areas of physics and mathematics.