Application of Fourier series to pressure waves

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SUMMARY

The discussion focuses on the application of Fourier series to analyze pressure waves, specifically a sawtooth function with a frequency of 1/2 Hz. The expression for the pressure change, ΔP(t), is derived from the Fourier expansion of the sawtooth function. It is concluded that only the oscillator resonating at 3/2 Hz will respond to the pressure wave, as the wave contains harmonics that are integral multiples of 1/2 Hz, while the 5/4 Hz oscillator does not match these harmonics.

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  • Understanding of Fourier series and expansions
  • Knowledge of pressure wave mechanics
  • Familiarity with harmonic frequencies and resonance
  • Basic calculus for manipulating equations
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  • Study the Fourier series representation of different waveforms
  • Learn about resonance in oscillatory systems
  • Explore the mathematical derivation of sawtooth functions
  • Investigate the effects of harmonic frequencies on resonators
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Sam Harrison
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Homework Statement



Assume that a pressure wave produces a change in pressure at a point in space [tex]\Delta P(t)[/tex] which is proportional to a sawtooth function of frequency f = 1/2 Hz.

(i) If the amplitude of the pressure wave is [tex]\Delta P_{0}[/tex], write down an expression for [tex]\Delta P(t)[/tex].

(ii) Two oscillators, designed to respond to changes in pressure, resonate at a frequency f = 3/2 Hz and f = 5/4 Hz respectively. When the pressure wave encounters the oscillators, which of these will resonate and why?

Homework Equations



The Fourier expansion of a sawtooth function [tex]f(x) = x, -1 < x < 1[/tex] is given by

[tex]f(x) = \sum_{r=1}^{\infty} \frac{-2(-1)^{r}}{\pi r} \sin{\pi r x}[/tex]

The Attempt at a Solution



Is the previous equation simply the answer to part (i) with the x's replaced with t's? Or is there a bit more to it? I think I'm missing something pretty obvious.

For part (ii), we know that [tex]r \pi t = 2 \pi ft[/tex] and hence [tex]r = 2f[/tex]. For f = 3/2 Hz, r = 3 and for f = 5/4 Hz, r = 5/2. Would only the first resonate as it is the only one that produces r that is an integer and hence satisfies the expansion above?
 
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Sam Harrison said:

Homework Statement



Assume that a pressure wave produces a change in pressure at a point in space [tex]\Delta P(t)[/tex] which is proportional to a sawtooth function of frequency f = 1/2 Hz.

(i) If the amplitude of the pressure wave is [tex]\Delta P_{0}[/tex], write down an expression for [tex]\Delta P(t)[/tex].

(ii) Two oscillators, designed to respond to changes in pressure, resonate at a frequency f = 3/2 Hz and f = 5/4 Hz respectively. When the pressure wave encounters the oscillators, which of these will resonate and why?

Homework Equations



The Fourier expansion of a sawtooth function [tex]f(x) = x, -1 < x < 1[/tex] is given by

[tex]f(x) = \sum_{r=1}^{\infty} \frac{-2(-1)^{r}}{\pi r} \sin{\pi r x}[/tex]


The Attempt at a Solution



Is the previous equation simply the answer to part (i) with the x's replaced with t's? Or is there a bit more to it? I think I'm missing something pretty obvious.
You have to get the amplitude right as well.
For part (ii), we know that [tex]r \pi t = 2 \pi ft[/tex] and hence [tex]r = 2f[/tex]. For f = 3/2 Hz, r = 3 and for f = 5/4 Hz, r = 5/2. Would only the first resonate as it is the only one that produces r that is an integer and hence satisfies the expansion above?
Yes. The sawtooth pressure wave contains harmonics with frequencies equal to integral multiples of 1/2 Hz, so it has a 3/2-Hz component but lacks a 5/4-Hz component and will excite only the one resonator.
 

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