Question

The period \(T\) (in seconds) of a pendulum is the time it takes for the pendulum to swing back and forth. The period is related to the length \(L\) (in inches) of the pendulum by the model \(T=2 \pi \sqrt{\frac{L}{384}} .\) Find the length of a pendulum with a period of eight seconds. Give your answer to the nearest tenth.

Short answer

The length \(L\) of the pendulum calculated using the given model equation should be approximately 390.2 inches, when rounded to the nearest tenth.

Step-by-step solution

Substitute the value of T in the equation

Given that the period \(T\) is 8 seconds, substitute 8 for \(T\) in the equation \(T=2 \pi \sqrt{\frac{L}{384}}\). This renders the equation as \(8 = 2 \pi \sqrt{\frac{L}{384}}\).

Manipulate the equation to solve for L

Re-arranging the equation to solve for the value of \(L\) essentially involves isolating \(L\) on one side of the equation by square both sides to eliminate the square root and dividing out \(4\pi^2\). This yields \(\frac{8^2}{(2\pi)^2}=\frac{L}{384}\). Multiply both sides of the equation by 384 to solve for \(L\). Doing so gives \(L=384\frac{8^2}{(2\pi)^2}.\)

Solve for L

Lastly, compute \(L=384\frac{8^2}{(2\pi)^2}\) using a calculator. The result will be the length of the Pendulum, which should then be rounded to the nearest tenth.

Key concepts

Mathematical Modeling

Understanding mathematical modeling is fundamental when attempting to decipher real-world problems that can be represented with mathematical expressions. A classic example of this is the period of a pendulum. Here, the relationship between the period, often denoted as \(T\), and the length of the pendulum, \(L\), is given by the model \(T=2 \pi \sqrt{\frac{L}{384}}\).

This formula encapsulates the physics behind the pendulum's motion, distilling it into a mathematical statement. For instance, it implies that the period of the pendulum's swing is proportional to the square root of its length. As a content creator aiming for ease of understanding, it’s crucial to ensure that students recognize the significance of modeling. It allows them to predict how changes in one variable (like the length of a pendulum) can affect another (the period of swing).

To translate this into a real-world scenario, imagine needing to design a clock with a pendulum that ticks every second. Using mathematical modeling, one can calculate the necessary length of the pendulum to achieve this period, thereby creating a clock that keeps time accurately. In essence, modeling is a bridge connecting abstract mathematics with tangible, observable phenomena.

Solving Algebraic Equations

Moving from the abstraction of mathematical modeling to the precision of solving algebraic equations, this next step is key to finding specific solutions to the problems at hand. Solving an algebraic equation involves finding the value(s) of the variable(s) that satisfy the equation. The problem provided centers around an equation where the variable \(L\) is hidden under a square root and multiplied by constants.

To make \(L\) the subject, one must manipulate the equation using algebraic operations. This process often involves isolating the variable on one side of the equation, which may include steps like squaring both sides to remove a square root, or performing addition, subtraction, multiplication, or division to both sides of the equation. These steps aim to maintain the equality while transforming the equation into a more solvable form. In the provided problem, the algebra involves squaring both sides to eliminate the square root, thereby paving the way to solve for \(L\) by subsequent multiplication and division. This highlights the importance of being fluent with algebraic manipulations to confidently tackle a variety of mathematical problems.

Square Roots in Algebra

The operation of taking a square root is frequently encountered in algebra and plays a pivotal role in our solution process. In the pendulum formula, the square root signifies the relationship between the period of the pendulum and its length; but in order to solve for \(L\), one must 'undo' this square root.

In algebra, squaring both sides of an equation where one side contains a square root effectively reverses the square root operation. As we do so, we must ensure to square all terms accordingly. For the pendulum period equation, when we square \(T=2 \pi \sqrt{\frac{L}{384}}\), we're left with \(T^2 = (2 \pi)^2 \left(\sqrt{\frac{L}{384}}\right)^2\), which simply becomes \(T^2 = 4\pi^2 \frac{L}{384}\) after simplifying. Mathematically, this elimination of the square root moves us one step closer to isolating \(L\).

Square roots can also introduce considerations for positive and negative solutions, since squaring a positive or negative number yields a positive result. However, in the context of physical problems like the period of a pendulum, only the positive root makes practical sense since we're often dealing with physical quantities that can't be negative, such as length. Therefore, when solving equations with square roots in real-world scenarios, it's as much about performing the right algebraic operations as it is about understanding the context of the problem.