Power and Amplitude of sound wave

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SUMMARY

The discussion focuses on calculating the amplitude of a longitudinal sound wave traveling through a copper rod using specific parameters such as frequency, radius, average power, Young's modulus, density, and wavelength. The correct formula for displacement amplitude A is A = Δp_o / (ωρc), where Δp_o represents pressure amplitude, ω is the angular frequency (2πf), ρ is the material density, and c is the speed of sound in the material, calculated as c = √(E/ρ). Additionally, the intensity of the sound wave is defined as I = P/a, with power P expressed as P = 1/2 ω²A²ρc.

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  • Understanding of wave mechanics and sound propagation
  • Familiarity with Young's modulus and material properties
  • Knowledge of angular frequency and its calculation
  • Basic principles of intensity and power in wave physics
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For my homework,
To find Amplitude of the wave,
for a longitudinal wave traveling down a copper rod.
Given: frequency, radius of copper rod, average power, Young's constant, density, wavelength.
what equation should I consider?
I'm thinking something like P=(omega*amplitud)^2 because I know power is proportional to amplitude and frequency of the wave...but I know that's wrong.
 
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The displacement amplitude A is given by:

A\,=\,\frac{\Delta{p_o}}{\omega\,\rho\,c}, where

\Delta{p_o} is the pressure amplitude,

\omega is the angular frequency given by 2\,\pi\,f,

\rho is the material density, and

c = speed of sound in the material, which is given by -

c = \sqrt{\frac{E}{\rho}}

where E is Young's (Elastic) modulus.


The intensity of the sound wave is I = P/a, where P is the power of the wave per unit transverse area, a, and

P = 1/2 \omega^2A2\rho c
 
Last edited:

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