Problem 95 - College Algebra and Calculus: An Applied Approach

Question

Period of a Pendulum The period \(T\) (in seconds) of a pendulum is given by \(T=2 \pi \sqrt{\frac{L}{32}}\) where \(L\) is the length (in feet) of the pendulum. Find the period of a pendulum whose length is 4 feet.

Step-by-Step Solution

Answer

The period of a pendulum of length 4 feet is approximately \( T \approx 2.22 \) seconds.

Step 1: Substitute Length into Formula

Firstly, note down the given length of the pendulum which is, \( L = 4 \) feet. Substitute this value into the formula \( T = 2 \pi \sqrt{\frac{L}{32}} \) to obtain \( T = 2 \pi \sqrt{\frac{4}{32}} \).

Step 2: Simplify under the square root

Next, simplify the fraction under the square root. The fraction \( \frac{4}{32} \) simplifies to \( \frac{1}{8} \). Substituting this value back into the formula we obtain \( T = 2 \pi \sqrt{\frac{1}{8}} \). Then we can simplify the square root to get \( T = 2 \pi * \frac{1}{2\sqrt{2}} \).

Step 3: Simplify the Expression

After cancellation of the redundant factors we get \( T = \frac{\pi}{\sqrt{2}} \). To obtain a numerical answer, substitute \( \pi \) by 3.14 and evaluate the square root of 2 using a calculator. After calculation we obtain \( T \approx 2.22 \) seconds.

Key Concepts

Pendulum Length

The length of a pendulum, denoted as \( L \), is a crucial parameter that influences its period, which is the time it takes to complete one oscillation. In physics, a pendulum consists of a weight suspended from a pivot, so it can swing freely. The length refers to the distance from the pivot point to the center of mass of the weight. Understanding this measurement allows us to use the period formula to calculate the pendulum’s movement. For instance, in the given exercise, our pendulum has a length of 4 feet.
This piece of information allows us to set up the initial substitution into the pendulum period formula.

Simplifying Square Roots

Simplifying square roots involves finding a simpler form of a square root expression, often by breaking down the expression into more manageable parts. In our problem, after substituting the length of the pendulum into the formula, we deal with the expression \( \sqrt{\frac{4}{32}} \). To simplify this, we start by simplifying the fraction: \( \frac{4}{32} \) is equivalent to \( \frac{1}{8} \).
This simplification makes it easier to handle the square root. Now, simplifying \( \sqrt{\frac{1}{8}} \) can further be written as \( \frac{1}{2\sqrt{2}} \). Mastery of such simplification techniques is vital in algebra as it lays groundwork for solving more complex expressions.

Algebraic Substitution

Algebraic substitution is a method where values are substituted into an expression or formula to simplify it. It’s a fundamental technique used to solve equations. In our context, we substitute the given pendulum length \( L = 4 \) into the period formula \( T = 2 \pi \sqrt{\frac{L}{32}} \).
After substitution, the expression becomes \( T = 2 \pi \sqrt{\frac{4}{32}} \). This step is essential as it converts the abstract equation into something tangible that we can calculate. By plugging in the values, we prepare the expression for further simplification and eventually for numerical evaluation.

Evaluating Numerical Expressions

Evaluating numerical expressions involves computing the final numerical result from a simplified expression. In the given problem, once we have simplified the square root part and have the expression \( T = \frac{\pi}{\sqrt{2}} \), we then evaluate it numerically.
Approximating \( \pi \) as 3.14 and using a calculator to estimate \( \sqrt{2} \) gives a precise numerical value for the period \( T \). Calculating gives \( T \approx 2.22 \) seconds. This process takes abstract mathematical results and makes them accessible and practical, like determining the time it takes for a pendulum to complete an oscillation.