Problem 62 - University Physics with Modern Physics

Question

A \(5.00-\mathrm{m}, 0.732-\mathrm{kg}\) wire is used to support two uni-? form \(235-\mathrm{N}\) posts of equal length (Fig. P15.62). Assume that the wire is essentially horizontal and that the speed of sound is 344 \(\mathrm{m} / \mathrm{s} . \mathrm{A}\) strong wind is blowing, causing the wire to vibrate in its 5 th overtone. What are the frequency and wavelength of the sound this wire produces?

Step-by-Step Solution

Answer

Frequency: 108.0 Hz, Wavelength: 3.19 m

Step 1: Calculate the Fundamental Frequency

The fundamental frequency \( f_1 \) of a vibrating string is given by:\[f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\]where \( L \) is the length of the wire, \( T \) is the tension in the wire, and \( \mu \) is the linear mass density.Given \( L = 5.00 \) m and \( \mu = \frac{0.732}{5.00} \ kg/m = 0.1464 \ kg/m \).To find \( T \), note that the tension in the wire is due to the weight of both posts, so \( T = 2 \times 235 \ N = 470 \ N \).Substituting these values gives:\[f_1 = \frac{1}{2 \times 5.00} \sqrt{\frac{470}{0.1464}} = \frac{1}{10} \sqrt{3210.5734} \approx 18.0 \ Hz\]

Step 2: Determine the Frequency of the 5th Overtone

The overtone frequency \( f_n \) is given by \( n \times f_1 \), where \( n \) is the mode number. For the 5th overtone, \( n = 6 \) because it is the 5th overtone (1 fundamental + 5 more overtones).\[f_6 = 6 \times 18.0 \ Hz = 108.0 \ Hz\]

Step 3: Calculate the Wavelength of the Sound Produced

The speed of sound \( v \) in air is given as 344 m/s. The wavelength \( \lambda \) of the sound wave can be determined using:\[\lambda = \frac{v}{f} = \frac{344}{108.0} \approx 3.19 \ m\]

Key Concepts

Vibrating String Frequency

When a string vibrates, it produces sound waves at specific frequencies, known as its natural frequencies. These frequencies are determined by three factors:

• The length of the string ( L ), which in our scenario is 5.00 meters
• The tension ( T ) in the string, here derived from the weight of two posts, resulting in 470 N
• The linear mass density ( μ ), calculated as the mass of the wire divided by its length, leading to 0.1464 kg/m in our example

Using these factors, we can compute the fundamental frequency, which is the lowest frequency at which the wire can vibrate. It's essential because it acts as the base for calculating higher overtones such as the 5th overtone discussed in the solution.
Whenever you encounter a problem about a vibrating string, always start by identifying these three factors. They offer a foundation for computing the string's frequencies.

Overtone Calculation

Overtones are frequencies that occur above the fundamental frequency. Each overtone represents a specific mode of vibration for the string.
The nth overtone is the (n+1)-th multiple of the fundamental frequency. For instance, the 5th overtone actually corresponds to the 6th multiple of the fundamental frequency.
In our problem, to find the frequency of the 5th overtone, we multiplied the fundamental frequency by 6:

• \[f_6 = n \times f_1 = 6 \times 18.0 \ Hz = 108.0 \ Hz\]

This approach efficiently determines higher frequencies the string will naturally support. Just keep in mind the correct formula for overtone calculation, and ensure you’ve accurately identified the fundamental frequency beforehand.

Wavelength Calculation

The wavelength of a sound wave is the distance between two consecutive points in phase, such as peaks or troughs.
It can be found using the formula:
\[\lambda = \frac{v}{f}\]
where \(v\) is the speed of sound, and \(f\) is the frequency of the generated sound. In the exercise, the speed of sound in air is provided as 344 m/s.
Given the overtone frequency, 108 Hz, the wavelength is calculated:

• \(\lambda = \frac{344}{108.0} \approx 3.19 \ m\)

This calculation allows us to understand how sound waves propagate from the vibrating string throughout the air. Wavelength is a critical aspect when exploring sound properties, especially in acoustics.

Fundamental Frequency

The fundamental frequency is the lowest natural frequency at which a system can vibrate. For a string, it's the frequency of the first harmonic, or the primary mode of vibration.
In the formula:
\[f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\]

• \(L\) is the length of the string, given in the problem as 5.00 m
• \(T\) is the tension, 470 N in this case, originating from supporting two 235 N posts
• \(\mu\) is the linear mass density, calculated here as 0.1464 kg/m

Plug these values into the formula to get the fundamental frequency, 18.0 Hz. Knowing this frequency is vital as it serves as a base multiple for calculating all overtones. The relation between harmonics and frequency is deeply rooted in physical principles affecting sound production in string instruments.

Linear Mass Density

Linear mass density, \(\mu\), is the mass per unit length of the wire. It's crucial because it affects how fast the wave travels through the string and, therefore, the frequency of sound it produces.
In the given problem, the linear mass density is determined by dividing the mass of the wire by its length:

• \(\mu = \frac{0.732\,kg}{5.00\,m} = 0.1464\,kg/m\)

Understanding linear mass density helps predict how variations in wire thickness or material might influence sound characteristics. It's one of the core components, along with tension and length, that dictate the fundamental frequency a string will produce. Always ensure to properly calculate \(\mu\) as it substantially impacts both tension computation and subsequent frequency analysis.