Problem 69 - Intermediate Algebra

Question

Find the period of a pendulum whose length is 2 feet. Give an exact answer and a two-decimal-place approximation.

Step-by-Step Solution

Answer

The period is approximately 1.57 seconds.

Step 1: Understanding the Pendulum Period Formula

The period of a simple pendulum can be calculated using 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. On Earth, \( g \approx 32.2 \frac{\text{ft}}{\text{s}^2} \).

Step 2: Plug in Known Values

Substitute the known values into the formula: \( L = 2 \) feet and \( g = 32.2 \frac{\text{ft}}{\text{s}^2} \). The equation becomes \( T = 2\pi \sqrt{\frac{2}{32.2}} \).

Step 3: Calculate the Inside of the Square Root

Compute \( \frac{2}{32.2} \). This value is approximately \( 0.06211 \).

Step 4: Compute the Square Root

Find the square root of \( 0.06211 \), which is approximately \( 0.2492 \).

Step 5: Multiply by \( 2\pi \)

Multiply the result from Step 4 by \( 2\pi \) to find the period: \( T \approx 2\pi \times 0.2492 \).

Step 6: Calculate the Exact and Approximate Period

The exact period is \( T = 2\pi \times 0.2492 \). Multiply to find the exact value: \( T \approx 1.5658 \) seconds. For a two-decimal-place approximation, \( T \approx 1.57 \) seconds.

Key Concepts

Pendulum Formula

The pendulum formula is a key equation for determining how long it takes a pendulum to swing back and forth once, known as its period. This formula is given by:\[T = 2\pi \sqrt{\frac{L}{g}} \]Here:

• \( T \) is the period of the pendulum, which we want to find.
• \( L \) represents the length of the pendulum, a crucial factor affecting the pendulum’s period.
• \( g \) is the acceleration due to gravity. On Earth, this is approximately \( 32.2 \text{ ft/s}^2 \) when using feet.

This formula assumes that the pendulum behaves ideally, with no air resistance and small oscillations.By applying this formula, we can predict how different lengths of pendulums will affect their oscillation time. It's a fundamental concept used in many physics problems.

Square Root

In the pendulum formula, the square root plays a significant role. The formula requires calculating the square root of \( \frac{L}{g} \).
For our exercise, we have:- \( L = 2 \) feet- \( g = 32.2 \text{ ft/s}^2 \)Plugging in these values:\[\frac{L}{g} = \frac{2}{32.2} \approx 0.06211\]Finding the square root of a number essentially means finding a value which, when multiplied by itself, gives the original number. Here, the square root of \( 0.06211 \) is approximately \( 0.2492 \).
This calculation is central to finding the pendulum's period, since it determines the multiplier of \( 2\pi \), a constant that finalizes the period.

Gravity Acceleration

Gravity acceleration, denoted as \( g \), is a measure of how quickly an object will accelerate when dropped. On Earth, in the context of pendulum calculations, \( g \) is roughly \( 32.2 \text{ ft/s}^2 \).
This value is essential in our formula, influencing how quickly the pendulum moves. It also ensures the calculation is tailored to the conditions on Earth's surface.
In different scenarios or planets, \( g \) would vary, significantly affecting your results.Gravity acceleration helps in:

• Determining how fast objects fall under gravity’s influence.
• Calculating the pendulum's period accurately on Earth.

Understanding \( g \)’s role in the pendulum formula clarifies why a pendulum's period isn't just dependent on its length, but also on the gravitational field it swings in.

Approximation Methods

Approximation methods are used when precise mathematical calculations lead to long, complex decimals. In our pendulum period problem, after computing \( 2\pi \times 0.2492 \), the result is approximately \( 1.5658 \).
To simplify this to two decimal places, it becomes \( 1.57 \). This technique makes the result more manageable and easier to communicate.
There are various reasons to approximate:

• It aids in quick estimations, which are particularly useful in homework or tests.
• Approximations can make complex numbers easier to understand and work with.

Using approximation methods doesn't mean the exact calculation isn’t important. It ensures the figures are suitable for everyday practical use, keeping things simple yet comprehensible.