Problem 10 - University Physics with Modern Physics

Question

(a) Show that the fractional change in the speed of sound \((d v / v)\) due to a very small temperature change \(d T\) is given by \(d v / v=\frac{1}{2} d T / T .\) (Hint: Start with Eq. \(16.10 .\) (b) The speed of sound in air at \(20^{\circ} \mathrm{C}\) is found to be 344 \(\mathrm{m} / \mathrm{s}\) . Use the result in part (a) to find the change in the speed of sound for a \(1.0^{\circ} \mathrm{C}\) change in air temperature.

Step-by-Step Solution

Answer

The speed of sound increases by approximately 0.59 m/s for a 1°C increase in air temperature.

Step 1: Understand the relationship

According to the equation for the speed of sound in terms of temperature, we have:\[ v = ext{constant} \times \sqrt{T} \]where \(v\) is the speed of sound and \(T\) is the absolute temperature in Kelvin. This means the speed of sound is directly proportional to the square root of the temperature.

Step 2: Differentiate the speed of sound equation

To find the fractional change in the speed of sound, we use calculus. Differentiate both sides with respect to \(T\):\[ \frac{dv}{dT} = \frac{1}{2} \times \text{constant} \times T^{-1/2} \]This differentiation gives us the rate of change of speed with respect to temperature.

Step 3: Express fractional change

Calculate the fractional change \(dv/v\). The original expression for \(v\) is:\[ v = \text{constant} \times T^{1/2} \]Differentiating gives:\[ \frac{dv}{v} = \frac{1}{2} \frac{dT}{T} \] This expression shows that the fractional change in speed is proportional to half the fractional change in temperature.

Step 4: Apply given temperature change

Given that \(v = 344\, \text{m/s}\) at \(20^\circ C\) and the temperature change \(dT = 1.0^\circ C\) or \(1\, \text{K}\), for small changes, the absolute temperature doesn’t significantly differ due to the increment. Thus, we use:\[ \frac{dv}{v} = \frac{1}{2} \cdot \frac{dT}{T} \]Substituting in the values gives:\[ \frac{dv}{344} = \frac{1}{2} \cdot \frac{1}{293} \] where 293 is the absolute temperature in Kelvin at \(20^\circ C\).

Step 5: Solve for speed change

From the above equation, calculate \(dv\):\[ dv = 344 \times \frac{1}{2} \times \frac{1}{293} \approx 344 \times 0.001707 \approx 0.587 \text{ m/s} \] Therefore, the change in speed of sound for a \(1.0^\circ C\) temperature change is approximately \(0.587\, \text{m/s}\).

Key Concepts

Temperature Dependence

The speed of sound in air is not constant; it varies with temperature. This happens because sound travels through the medium's particles, and their movement changes with temperature. As the air heats up, the particles move faster, allowing sound to travel quicker.
This relationship is mathematically expressed by the equation:

• \[ v = ext{constant} \times \sqrt{T} \]

where \(v\) denotes the speed of sound and \(T\) is the absolute temperature in Kelvin. This equation indicates that the speed of sound is directly proportional to the square root of the temperature. Therefore, even a small change in temperature can lead to a noticeable change in the speed of sound due to its temperature dependence.

Fractional Change

Understanding fractional change involves looking at how a small increase or decrease in temperature can cause a proportional change in the speed of sound. In the context of sound speed, we denote fractional change as \(\frac{dv}{v}\), which measures how the speed of sound varies relative to its current value. Calculus is used in deriving this relation

• First, we differentiate the speed equation: \( v = ext{constant} \times \sqrt{T} \)
• On differentiating, we find \( \frac{dv}{dT} = \frac{1}{2} \text{constant} \times T^{-1/2} \)

This results in the fractional change being expressed as:

• \[ \frac{dv}{v} = \frac{1}{2} \frac{dT}{T} \]

Thus, the fractional change in speed of sound is proportional to half the fractional change in temperature. This means a small temperature increment leads to a predictable and proportional increase in sound speed.

Calculus in Physics

Calculus plays a crucial role in understanding the dynamics of physical systems, such as how variables like temperature and speed of sound interact. By using differentiation, it helps in determining the rate of change of one variable with respect to another. In our problem, we employ calculus to:

• Determine how the speed of sound varies with a change in temperature
• Express this change as \( \frac{dv}{v} = \frac{1}{2} \frac{dT}{T} \)

When we perform differentiation on physical relationships, we obtain useful insights into the nature of these relationships. It helps us predict how minor changes in initial conditions, like temperature in this instance, affect other properties like speed. Thus, calculus allows scientists and students alike to frame and solve complex physics problems with precision and clarity.