Problem 80 - Precalculus: Functions and Graphs

Question

Speed of sound The speed of sound in air at \(0^{\circ} \mathrm{C}(\text { or } 273 \mathrm{K})\) is \(1087 \mathrm{ft} / \mathrm{sec},\) but this speed increases as the temperature rises. The speed \(v\) of sound at temperature \(T\) in \(K\) is given by \(v=1087 \sqrt{T / 273} .\) At what temperatures does the speed of sound exceed \(1100 \mathrm{ft} / \mathrm{sec} ?\)

Step-by-Step Solution

Answer

The speed of sound exceeds 1100 ft/sec at temperatures above approximately 279 K.

Step 1: Set up the equation

We are given the formula for the speed of sound as \(v = 1087 \sqrt{\frac{T}{273}} \). We need to find the temperature at which \(v > 1100 \text{ ft/sec}\). To do this, set up the inequality: \(1087 \sqrt{\frac{T}{273}} > 1100\).

Step 2: Isolate the square root

Divide both sides of the inequality by 1087 to isolate the square root: \(\sqrt{\frac{T}{273}} > \frac{1100}{1087}\).

Step 3: Remove the square root

Square both sides of the inequality to remove the square root: \(\frac{T}{273} > \left(\frac{1100}{1087}\right)^2\).

Step 4: Solve for T

Multiply both sides by 273 to solve for \(T\): \(T > 273 \times \left(\frac{1100}{1087}\right)^2\).

Step 5: Calculate the numerical value

Calculate \(\left(\frac{1100}{1087}\right)^2\) which is approximately \(1.0236\). Then, \(T > 273 \times 1.0236\). Calculate the product to find \(T\), which is approximately \(279.46\).

Key Concepts

Speed of Sound

The speed of sound is a measure of how quickly sound waves travel through a medium like air, water, or solids. In general, sound moves faster in warmer conditions. This is because higher temperatures provide more energy to the particles in a medium, making them vibrate more quickly. At those higher energy levels, sound waves can move more easily through the medium.

In air, the speed of sound at a standard temperature of \(0^{\circ} \mathrm{C}\) or \(273 \mathrm{K}\) is approximately \(1087 \mathrm{ft/sec}\). This speed can be calculated using the formula:

• \(v = 1087 \sqrt{\frac{T}{273}}\)

where \(v\) represents the speed of sound and \(T\) is the temperature in Kelvin. This formula shows how speed increases with temperature due to the \(\sqrt{T/273}\) term, which results in a larger value for \(T > 273\).

Inequality Solving

Inequality solving is a crucial part of precalculus and arithmetic. It is used to find the range of values for variables that satisfy expressions involving inequalities.In the context of our problem, we are solving the inequality:

• \(1087 \sqrt{\frac{T}{273}} > 1100\)

The solution is to isolate \(T\) by systematically manipulating the expression. We:

• Divide both sides by \(1087\)
• Square both sides to eliminate the square root
• Multiply by \(273\) to solve for \(T\)

This process shows how inequalities are not only used in sound calculations but in various practical applications, like physics and engineering.

It reinforces logical problem-solving and helps students develop critical thinking skills.

Temperature

Temperature is a measure of the average kinetic energy of the particles in a substance. It's typically measured in degrees Celsius, Kelvin, or Fahrenheit. In scientific contexts, Kelvin (K) is frequently used since it is an absolute scale starting at absolute zero, where particles have minimal motion.

Understanding temperature is vital as it affects various physical phenomena, like the speed of sound. As seen in the formula \(v = 1087 \sqrt{\frac{T}{273}}\), the temperature \(T\) directly affects the calculation of sound speed in air.

Solving the exercise also shows the difference between temperatures as understood in different scales, reminding students of the conversion need, for instance, \(0^{\circ} \mathrm{C}\) is \(273 \mathrm{K}\). Thus, the exercise helps correlate mathematical expressions with real-world units and measures.

Sound Waves

Sound waves are vibrations that travel through the air or other media. These waves are mechanical, meaning they require a medium (like air or water) to travel through.

The speed of sound depends on factors like temperature, humidity, and the medium itself (for example, faster in solids than gases).

Sound travels as a longitudinal wave, where the displacement of the medium is in the same direction as the wave. Understanding sound as a concept involves recognizing how these waves interact with different environments and how variables—like temperature—affect their speed. This is vital for applications ranging from meteorology to music production.

By using mathematics to calculate the speed of sound, students blend theoretical knowledge about wave mechanics with practical, real-world problem-solving.