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Chapter 15: Problem 22

A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplitude 1.6 mm travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

Short Answer

Expert verified
(a) 0.928 W; (b) 0.232 W when amplitude is halved.

Step by step solution

01

Calculate the wave speed

First, determine the speed of the wave on the wire using the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T = 25.0 \) N is the tension and \( \mu = \frac{m}{L} \) is the linear mass density. Here, \( m = 3.00 \, \text{g} = 0.0030 \, \text{kg} \) and \( L = 80.0 \, \text{cm} = 0.80 \, \text{m} \). First, calculate \( \mu = \frac{0.0030}{0.80} = 0.00375 \, \text{kg/m} \). Substitute \( \mu \) into the wave speed formula: \( v = \sqrt{\frac{25}{0.00375}} = \sqrt{6666.67} \approx 81.65 \, \text{m/s} \).
02

Calculate the average power of the wave

The average power \( P \) carried by a wave on a stretched string can be calculated using the formula \( P = \frac{1}{2} \mu v \omega^2 A^2 \), where \( \omega \) is the angular frequency and \( A \) is the amplitude. First, find \( \omega = 2\pi f = 2\pi (120) = 240\pi \, \text{rad/s} \). Convert amplitude \( A \) to meters: \( A = 1.6 \, \text{mm} = 0.0016 \, \text{m} \). Substituting all values: \[ P = \frac{1}{2} \times 0.00375 \times 81.65 \times (240\pi)^2 \times (0.0016)^2 \approx 0.928 \, \text{W} \].
03

Analyze effect of halving the amplitude

If the wave amplitude is halved, \( A \) becomes \( 0.0008 \, \text{m} \). Since power \( P \) is proportional to \( A^2 \), the new power can be calculated as follows: Halving the amplitude means the new power \( P' \) is \( (0.0008/0.0016)^2 \times 0.928 \, \text{W} = \frac{1}{4} \times 0.928 \, \text{W} = 0.232 \, \text{W} \).
04

Conclusion: Final Answer

The average power carried by the wave is approximately 0.928 W. If the amplitude is halved, the average power becomes 0.232 W.

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Key Concepts

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

wave speed
Wave speed is an essential concept in understanding waves on a string. It refers to how fast a wave travels along the string, which depends largely on the tension in the string and the linear mass density. The formula for the wave speed (\(v\)) on a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Here, \(T\) is the tension in the string in newtons (N), and \(\mu\) is the linear mass density in kilograms per meter (kg/m).
  • The higher the tension, the faster the wave travels.
  • Conversely, a larger mass density, meaning a heavier string, results in a slower wave speed.
In practical terms, knowing the wave speed helps us understand how quickly the energy is being transferred along the string, which is crucial for calculating wave power.
tension in a string
The tension in a string is the force exerted along the string that keeps it stretched. Tension is a key factor that impacts both wave speed and the overall dynamics of the wave. In the given problem, the piano wire is stretched with a tension of 25.0 N. This force is necessary to maintain the wire at a certain tightness. Tension affects:
  • Wave Speed: As previously mentioned, higher tension means higher wave speed.
  • Wave Frequency: While more indirectly, tension can also influence the frequency at which the string vibrates.
Tension is crucial in musical instruments like the piano since it affects the note played鈥攖he tighter the string, the higher the pitch.
linear mass density
Linear mass density (\(\mu\)) is a measure that combines the mass and length of the string into a single value. It indicates how much mass is distributed along each unit length of the string. For the piano wire in the example, with a mass of 3.00 g and a length of 80.0 cm, the linear mass density is calculated as:\[ \mu = \frac{\text{mass}}{\text{length}} = \frac{0.0030 \, \text{kg}}{0.80 \, \text{m}} = 0.00375 \, \text{kg/m} \]Understanding linear mass density is vital for two main reasons:
  • It directly affects the wave speed; a higher \(\mu\) means a slower wave.
  • It is essential for determining the wave's transport of energy, as seen in the power formula.
When designing strings for specific purposes like music or communication cables, considering \(\mu\) ensures the desired performance.
wave amplitude
Wave amplitude is the maximum displacement from the rest position that the particles of the medium experience as the wave passes. The exercise provides an amplitude of 1.6 mm (or 0.0016 m) for the wave traveling along the piano wire. Amplitude plays a significant role in determining the power that a wave transfers:
  • The wave's energy is directly proportional to the square of its amplitude, which is why halving the amplitude to 0.0008 m results in the power being reduced to a quarter.
  • Simply put, a larger amplitude means a more energetic wave, leading to higher power.
Understanding how amplitude affects power can help in applications where controlling the energy transfer is crucial. For instance, in musical contexts, a larger amplitude typically means a louder sound.

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Most popular questions from this chapter

A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 g. (a) What is the frequency of its fundamental mode of vibration? (b) What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to 10,000 Hz?

You are investigating the report of a UFO landing in an isolated portion of New Mexico, and you encounter a strange object that is radiating sound waves uniformly in all directions. Assume that the sound comes from a point source and that you can ignore reflections. You are slowly walking toward the source. When you are 7.5 m from it, you measure its intensity to be 0.11 W/m\(^2\). An intensity of 1.0 W/m\(^2\) is often used as the "threshold of pain." How much closer to the source can you move before the sound intensity reaches this threshold?

A water wave traveling in a straight line on a lake is described by the equation $$y(x, t) = (\mathrm{2.75 \, cm) cos(0.410 \mathrm{rad/cm} \, \textit{x}} + 6.20 \, \mathrm{rad}/s \; t)$$ where y is the displacement perpendicular to the undisturbed surface of the lake. (a) How much time does it take for one complete wave pattern to go past a fisherman in a boat at anchor, and what horizontal distance does the wave crest travel in that time? (b) What are the wave number and the number of waves per second that pass the fisherman? (c) How fast does a wave crest travel past the fisherman, and what is the maximum speed of his cork floater as the wave causes it to bob up and down?

A light wire is tightly stretched with tension F. Transverse traveling waves of amplitude \(A\) and wavelength \(\lambda_1\) carry average power \(P_{av,1} = 0.400\) W. If the wavelength of the waves is doubled, so \(\lambda_2 = 2\lambda_1\), while the tension \(F\) and amplitude \(A\) are not altered, what then is the average power \(P_{av,2}\) carried by the waves?

You are exploring a newly discovered planet. The radius of the planet is \(7.20 \times 10^7\) m. You suspend a lead weight from the lower end of a light string that is 4.00 m long and has mass 0.0280 kg. You measure that it takes 0.0685 s for a transverse pulse to travel from the lower end to the upper end of the string. On the earth, for the same string and lead weight, it takes 0.0390 s for a transverse pulse to travel the length of the string. The weight of the string is small enough that you ignore its effect on the tension in the string. Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?
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