Problem 28
Question
To review the concepts that play roles in this problem, consult Multiple- Concept Example 4. Sometimes, when the wind blows across a long wire, a low- frequency “moaning" sound is produced. This sound arises because a standing wave is set up on the wire, like a standing wave on a guitar string. Assume that a wire (linear density \(=0.0140\) \(\mathrm{kg} / \mathrm{m}\) ) sustains a tension of \(323 \mathrm{~N}\) because the wire is stretched between two poles that are \(7.60 \mathrm{~m}\) apart. The lowest frequency that an average, healthy human ear can detect is \(20.0 \mathrm{~Hz} .\) What is the lowest harmonic number \(n\) that could be responsible for the “moaning" sound?
Step-by-Step Solution
Verified Answer
The lowest harmonic number responsible for the moaning sound is 2.
1Step 1: Identify Given Values
From the problem statement, we have the following given values: linear density \( \mu = 0.0140 \; \text{kg/m} \), tension \( T = 323 \; \text{N} \), length of the wire \( L = 7.60 \; \text{m} \), and the lowest frequency \( f = 20.0 \; \text{Hz} \).
2Step 2: Use the Wave Speed Formula
The speed of a wave on the wire can be calculated using the formula for wave speed on a string: \( v = \sqrt{\frac{T}{\mu}} \). Substitute the given tension and linear density into this formula to find the wave speed.
3Step 3: Calculating Wave Speed
Substitute the known values: \( v = \sqrt{\frac{323 \; \text{N}}{0.0140 \; \text{kg/m}}} = \sqrt{23,071.43} \approx 151.8 \; \text{m/s} \).
4Step 4: Use the Harmonic Frequency Formula
The frequency of a standing wave in a wire is given by the formula \( f = \frac{n \cdot v}{2L} \), where \( n \) is the harmonic number. Rearrange this to solve for \( n \): \( n = \frac{2L \cdot f}{v} \).
5Step 5: Calculating the Harmonic Number
Substitute the values into the adjusted formula: \( n = \frac{2 \cdot 7.60 \cdot 20.0}{151.8} \approx \frac{304}{151.8} \approx 2.0026 \). Since \( n \) must be a whole number, round to the nearest whole number to get \( n = 2 \).
Key Concepts
Wave SpeedHarmonic FrequencyLinear DensityTension in a Wire
Wave Speed
Wave speed is an essential concept in understanding how waves travel through different mediums. It tells us how fast a wave is moving, which is crucial for solving problems related to standing waves. For waves on a string, like a wire, the wave speed (\(v\)) can be determined through the formula:\[v = \sqrt{\frac{T}{\mu}}\]where:
- \(T\) is the tension applied to the string or wire.
- \(\mu\) is the linear density, representing how much mass is in a given length of the wire.
Harmonic Frequency
Harmonic frequency refers to the different frequencies that can occur on a string or wire, resulting in standing waves. Each frequency corresponds to a harmonic, which is characterized by a particular pattern or mode of vibration. To find the frequency of a particular harmonic, the formula is:\[f = \frac{n \cdot v}{2L}\]where:
- \(f\) is the frequency of the harmonic.
- \(n\) is the harmonic number, indicating the order of the mode (1st harmonic, 2nd harmonic, etc.).
- \(v\) is the wave speed.
- \(L\) is the length of the wire.
Linear Density
Linear density is an important concept when dealing with wave mechanics on wires and strings. It describes how much mass is distributed over the length of the wire. Essentially, it's a measure of the "heaviness" of the wire per meter, given by the formula:\[\mu = \frac{m}{L}\]where:
- \(\mu\) is the linear density in kilograms per meter (kg/m).
- \(m\) is the mass of the wire.
- \(L\) is the length of the wire.
Tension in a Wire
Tension is the force exerted along a wire or string. This force stretches the wire and influences the characteristics of any waves that travel through it. By altering the tension, we can change the speed and frequency of waves on the wire.The formula linking tension to wave speed is:\[v = \sqrt{\frac{T}{\mu}}\]where:
- \(T\) is tension, measured in Newtons (N).
- \(\mu\) is the linear density of the wire.
Other exercises in this chapter
Problem 26
Multiple-Concept Example 4 deals with the same concepts as this problem. A \(41-\mathrm{cm}\) length of wire has a mass of \(6.0 \mathrm{~g}\). It is stretched
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