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Chapter 16: Problem 10

(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.

Short Answer

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

Step by step solution

01

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.
02

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.
03

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.
04

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鈥檛 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 \(20^\circ C\).
05

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 \(0.587\, \text{m/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

(a) What is the sound intensity level in a car when the sound intensity is 0.500\(\mu \mathrm{W} / \mathrm{m}^{2} 7\) (b) What is the sound intensity level in the air near a jackhammer when the pressure amplitude of the sound is 0.150 \(\mathrm{Pa}\) and the temperature is \(20.0^{\circ} \mathrm{C}\) ?

Two identical taut strings under the same tension \(F\) produce a note of the same fundamental frequency \(f_{0}\) . The tension in one of them is now increased by a very small amount \(\Delta F\) . (a) If they are played together in their fundamental, show that the frequency of the beat produced is \(f_{\text { keet }}=f_{0}(\Delta F / 2 F)\) . (b) Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 \(\mathrm{Hz}\) . One of the strings is retuned by increasing its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously at their centers. By what percentage was the string tension changed?

You blow across the open mouth of an empty test tube and produce the fundamental standing wave of the air column inside the test tube. The speed of sound in air is 344 \(\mathrm{m} / \mathrm{s}\) and the test tube acts as a stopped pipe. (a) If the length of the air column in the test tube is 14.0 \(\mathrm{cm}\) , what is the frequency of this standing wave? (b) What is the frequency of the nfundamental standing wave in the air column if the test tube is half filled with water?

How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we recelve from it is 10.0\(\%\) higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)

Doppler Radar. A giant thunderstorm is moving toward a weather station at \(45.0 \mathrm{mi} / \mathrm{h}(20.1 \mathrm{m} / \mathrm{s}) .\) If the station sends a radar beam of frequency 200.0 \(\mathrm{MHz}\) toward the storm, what is the difference in frequency between the emitted beam and the beam reflected back from the storm? Be careful to carry plenty of significant figures! (Hint The storm reflects the same frequency that it receives.)
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