Introductory Simple Harmonic Motion Problem

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SUMMARY

The discussion focuses on calculating the maximum speed of a tympanic membrane oscillation when subjected to a sound wave with a frequency of 600 Hz and an amplitude of 1.0 nm. The formula used is Vmax = ωA, where ω is the angular velocity calculated as ω = 2πf. The correct approach involves substituting the frequency into the equation to find the maximum speed, confirming that the method of algebraic manipulation before numerical substitution is best practice.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with the relationship between frequency and angular velocity
  • Basic algebra skills for manipulating equations
  • Knowledge of sound wave properties and their effects on physical systems
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  • Study the derivation of the formula Vmax = ωA in detail
  • Explore the implications of amplitude and frequency on wave behavior
  • Learn about the physical properties of tympanic membranes in response to sound
  • Investigate other applications of simple harmonic motion in acoustics
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Students studying physics, particularly those focusing on acoustics and wave motion, as well as educators looking for clear examples of simple harmonic motion calculations.

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Q: When a sound wave with a certain intensity is detected by the tympanic membrane, the amplitude of the resultant motion is 1.0 nm (1.0 x 10-9 m). If the frequency of the sound is 600 Hz, what is the maximum speed of the membrane oscillation?

My answer:

1)Vmax = wA (where w= angular velocity)
2)w = 2(pi) x f = 2(pi) x 600 Hz
3) ^Then I insert the above result of 'w' into the first equation

Is this correct? Additionally, is there a simpler way to do this?

Any replies will be very much appreciated :)
 
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That's pretty much it yep - multiplying a bunch of numbers together is as simple as it gets.

It is best practice to do the algebra before you plug in the numbers though ... so for oscillations of amplitude A and frequency f, the max speed for the membrane is ##v = \omega A = 2\pi f A##.
 
That seems simple enough, I'd think.
 

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