Problem 113 - College Algebra with Modeling and Visualization

Question

The frequency \(F\) of \(a\) vibrating string is directly proportional to the square root of the tension \(T\) on the string and inversely proportional to the length \(L\) of the string. If both the tension and the length are doubled, what happens to \(F ?\)

Step-by-Step Solution

Answer

The frequency is multiplied by \( \frac{\sqrt{2}}{2} \).

Step 1: Understand the Relationship

The problem states that the frequency \( F \) is directly proportional to the square root of the tension \( T \), and inversely proportional to the length \( L \). This can be expressed as: \[ F \propto \frac{\sqrt{T}}{L}. \] We can write this relationship as an equation using a constant of proportionality \( k \): \[ F = k \frac{\sqrt{T}}{L}. \]

Step 2: Initial Setup of the Equation

Given the equation \( F = k \frac{\sqrt{T}}{L} \), we need to understand what happens if both \( T \) and \( L \) are doubled. Initially, the equation can be written as: \[ F = k \frac{\sqrt{T_1}}{L_1}, \] where \( T_1 \) and \( L_1 \) are the initial tension and length, respectively.

Step 3: Modify for Doubled Tension and Length

When both the tension and the length are doubled, we have new values \( T_2 = 2T_1 \) and \( L_2 = 2L_1 \). Substitute these into the equation: \[ F' = k \frac{\sqrt{2T_1}}{2L_1}. \]

Step 4: Simplify the Equation

Simplify \( F' \) by isolating terms: \[ F' = k \frac{\sqrt{2} \cdot \sqrt{T_1}}{2L_1} = \frac{k\sqrt{2}}{2} \cdot \frac{\sqrt{T_1}}{L_1}. \]

Step 5: Compare with Original Frequency

The original frequency is \( F = k \frac{\sqrt{T_1}}{L_1} \) and the new frequency is \( F' = \frac{\sqrt{2}}{2} F \). Thus, the new frequency \( F' \) is different by a factor of \( \frac{\sqrt{2}}{2} \) compared to its original value.

Key Concepts

Frequency of Vibrating String

When we talk about the frequency of a vibrating string, we are referring to how many times per second the string vibrates. This frequency determines the pitch of the sound produced by the string. If the string vibrates at a higher frequency, it produces a higher pitch, while a lower frequency results in a lower pitch. The frequency is influenced by several factors:

• Tension: Increasing the tension generally increases the frequency, producing a higher pitch.
• Length: A longer string tends to vibrate at a lower frequency.
• Mass density: This also affects how the string vibrates, though it’s not directly considered in this particular exercise.

Understanding how these factors come together to affect frequency is vital to predicting how the string will behave under different conditions.

Direct and Inverse Proportionality

Direct and inverse proportionality are two fundamental concepts in understanding how different variables relate to each other in mathematical modeling.

• Direct Proportionality: This means that as one variable increases, the other variable also increases. In the problem, the frequency \( F \) is directly proportional to the square root of the tension \( T \). If the tension increases, the frequency does too.
• Inverse Proportionality: In this scenario, as one variable increases, the other decreases. For this exercise, frequency \( F \) is inversely proportional to the length \( L \) of the string. Therefore, increasing the length of the string decreases the frequency.

These proportional relationships help us understand how altering tension and length affects the vibration frequency of the string.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations and solving for variables, which is a crucial skill in mathematical modeling. In the given exercise, we rearrange the initial relationship \( F = k \frac{\sqrt{T}}{L} \) to see the effect of changing tension and length. Here are the basic steps:

• Substitute initial values \( T_1 \) and \( L_1 \) into \( F = k \frac{\sqrt{T_1}}{L_1} \).
• Adjust formula for new values \( T_2 = 2T_1 \) and \( L_2 = 2L_1 \) into \( F' = k \frac{\sqrt{2T_1}}{2L_1} \).
• Simplify the equation to find \( F' = \frac{k\sqrt{2}}{2} \frac{\sqrt{T_1}}{L_1} \).

Through manipulation, we understand how variables interact and how changes impact frequency.

Square Roots in Equations

Square roots play an essential role in this problem, especially when analyzing frequency's dependence on tension. The original problem establishes that frequency is proportional to \( \sqrt{T} \). Key aspects include:

• Square roots express the idea that frequency changes at a rate slower than tension. Double the tension does not double the frequency, it changes by \( \sqrt{2} \).
• When we expanded the relationship with \( T_2 = 2T_1 \), the square root of the tension becomes \( \sqrt{2T_1} = \sqrt{2}\cdot\sqrt{T_1} \).
• This highlights how Square roots can both complicate and deepen our understanding of relationships in equations.

Understanding square roots helps determine how more nuanced variable relationships work, especially in physical models.