Problem 4 - University Physics with Modern Physics

Question

A loud factory machine produces sound having a displace ment amplitude of \(1.00 \mu \mathrm{m},\) but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 \(\mathrm{Pa.}\) Under the conditions of this factory, the bulk modulus of air is \(1.42 \times 10^{5}\) Pa. What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit? Is this frequency audible to the workers?

Step-by-Step Solution

Answer

The highest frequency is approximately 769.23 Hz, which is audible.

Step 1: Understanding the Problem

We need to find the maximum frequency of sound that does not exceed a given pressure amplitude of 10.0 Pa. Given variables include: displacement amplitude, maximum pressure amplitude, and bulk modulus. Use the relation between pressure amplitude, displacement amplitude, frequency, and bulk modulus to solve this problem.

Step 1: Formula for Pressure Amplitude

The pressure amplitude \( P_0 \) can be expressed in terms of displacement amplitude \( A \), frequency \( f \), and the bulk modulus \( B \). The formula is: \[ P_0 = B k A \] where \( k \) is the wave number which is equal to \( 2\pi f/v \) and \( v \) is the speed of sound.

Step 2: Substitute Known Values

To use the formula \( P_0 = B k A \), we substitute \( k = \frac{2\pi f}{v} \) into the equation: \[ P_0 = B \left( \frac{2 \pi f}{v} \right) A \]Simplify it to: \[ P_0 = \frac{2 \pi f B A}{v} \] We have \( B = 1.42 \times 10^5 \: \text{Pa} \) and \( A = 1.00 \times 10^{-6} \: \text{m} \).

Step 3: Calculate the Speed of Sound

The speed of sound \( v \) in air can be approximately taken as \( 343 \: \text{m/s} \) at room temperature, assuming standard conditions inside the factory.

Step 4: Solve for Frequency

Rearrange the equation to solve for frequency \( f \): \[ f = \frac{P_0 v}{2 \pi B A} \] Substitute the given values to find \( f \): \[ f = \frac{10 \times 343}{2 \pi \times 1.42 \times 10^5 \times 1 \times 10^{-6}} \] Calculate the result.

Step 5: Determine Audibility

Calculate the frequency and check its range. The human audible range is approximately 20 Hz to 20,000 Hz. Check whether the calculated frequency lies within this range.

Key Concepts

Displacement Amplitude

The displacement amplitude of a sound wave refers to how much the particles of the medium move from their rest position as the wave passes through. In this problem, the displacement amplitude is given as 1.00 micrometer, or \(1.00 \times 10^{-6} \) meters. This small movement is sufficient to create a perceivable sound under the conditions of the factory.

The displacement amplitude is directly related to the intensity of the sound: the larger the amplitude, the more intense the sound. In this scenario, our main interest is in connecting this amplitude with pressure changes, which leads to the concept of pressure amplitude.

This linkage is what allows us to use the displacement amplitude to calculate other parameters of the wave, such as frequency and pressure, using relevant physical equations.

Pressure Amplitude

Pressure amplitude is an important concept in sound waves, which measures the maximum change in pressure as the wave passes through a medium. The problem specifies that the maximum allowable pressure amplitude is 10.0 Pa to prevent hearing damage to workers.

It connects directly to both displacement amplitude and frequency through the equation: - \( P_0 = \frac{2 \pi f B A}{v} \)
This formula includes the bulk modulus \(B\), displacement amplitude \(A\), speed of sound \(v\), and frequency \(f\). Understanding this equation is key to solving the exercise, as it helps to find the maximum frequency of the machine's sound under the given pressure restriction.

Whenever dealing with sound waves, pressure amplitude is crucial, as it helps us feel the sounds around us and can explain why, at some levels, sounds can become dangerous to hearing.

Bulk Modulus

Bulk modulus is a measure of a substance's resistance to uniform compression. In the context of sound waves, it affects how waves travel through the medium. The higher the bulk modulus, the harder it is to compress the medium, hence affecting the speed of sound and pressure changes.

In this exercise, the bulk modulus of air is provided as \(1.42 \times 10^5\) Pa. This value plays an essential role in determining how the pressure amplitude relates to the displacement amplitude and frequency.
The bulk modulus is part of the pressure amplitude formula, \(P_0 = \frac{2\pi f B A}{v}\), indicating how intrinsic properties of the material or medium, like its compressibility, influence sound wave behavior.

Frequency Calculation

To ensure the factory machine's sound is not harmful to workers, we need to calculate and limit the frequency of sound waves. Frequency, which is the number of complete oscillations per second, expressed in Hertz (Hz), determines the sound's pitch and is related to the wave's characteristics.

Using the rearranged equation: - \( f = \frac{P_0 v}{2\pi BA} \)
We substitute given values, including the pressure amplitude, speed of sound, bulk modulus, and displacement amplitude.
This calculation helps us find the maximum frequency without exceeding the 10.0 Pa pressure amplitude limit. The computed frequency is then compared with human audible range, confirming whether it is within the human audible frequency range of 20 Hz to 20,000 Hz, which is crucial to ensure the sound stays safe and detectable.