Calculating Diaphragm Acceleration in a Loudspeaker

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Homework Help Overview

The discussion revolves around calculating the frequency at which the acceleration of a loudspeaker diaphragm equals the acceleration due to gravity (g). The problem involves understanding the oscillation of the diaphragm with a specified amplitude of 0.9 µm and the implications of the cosine term in the acceleration equation.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the validity of maximizing the cosine term in the acceleration equation, questioning whether the assumption of perfect vibration is justified without explicit information on the diaphragm's construction. Some express concern about the clarity of the problem statement, suggesting it may not adequately specify the conditions under which the maximum acceleration occurs.

Discussion Status

The discussion is ongoing, with participants offering insights into the assumptions made in the problem. There is recognition of potential ambiguities in the problem's wording, particularly regarding the maximum amplitude and the conditions for achieving maximum acceleration. No consensus has been reached, but various interpretations are being explored.

Contextual Notes

Participants note that the original problem may have been poorly written, as it does not clearly specify whether it is asking for maximum acceleration at maximum amplitude or if other conditions might apply. The textbook reference adds context but also raises further questions about the interpretation of acceleration at different frequencies.

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Homework Statement



A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to 0.9 µm.
At what frequency is the magnitude a of the diaphragm's acceleration equal to g?

Homework Equations



[tex]a=-\omega^2 x_m \cos(\omega t +\phi)[/tex]

The Attempt at a Solution



I decided to chuck the cosine term overboard to see what happened... and I got the right answer (525.45 Hz)! Why am I allowed to maximize the cosine term if I'm not told that the diaphragm is constructed to maximize efficiency?
 
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Why am I allowed to maximize the cosine term if I'm not told that the diaphragm is constructed to maximize efficiency?
It's safe to say that the perfect vibration of the loudspeaker was an expected assumption. Otherwise they would have to give you more information.

If there's nothing acting against the vibration then the cosine term will always fluctuate from -1 to 1, and nothing else will change. Because the cosine is the only varying factor, and we have assumed a perfect loudspeaker, we can set it to unity.
 
mbrmbrg said:

Homework Statement



A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to 0.9 µm.
At what frequency is the magnitude a of the diaphragm's acceleration equal to g?

Homework Equations



[tex]a=-\omega^2 x_m \cos(\omega t +\phi)[/tex]

The Attempt at a Solution



I decided to chuck the cosine term overboard to see what happened... and I got the right answer (525.45 Hz)! Why am I allowed to maximize the cosine term if I'm not told that the diaphragm is constructed to maximize efficiency?

If the question has been posted correctly, it is poorly written. It should be asking for the frequency at which the maximum acceleration of the diaphragm is g when the speaker is driven at its maximum permitted amplitude. Higher frequencies can cause that same acceleration at some point in the motion of the speaker, and g could be the maximum acceleration achieved at those higher frequencies if the speaker is driven at less than maximum amplitude. Setting the cosine to 1 and using x_m in the calculation is assuming g is the maximumum acceleration and that it occurs at that maximum permitted amplitude.
 
OlderDan said:
If the question has been posted correctly, it is poorly written. It should be asking for the frequency at which the maximum acceleration of the diaphragm is g when the speaker is driven at its maximum permitted amplitude. Higher frequencies can cause that same acceleration at some point in the motion of the speaker, and g could be the maximum acceleration achieved at those higher frequencies if the speaker is driven at less than maximum amplitude. Setting the cosine to 1 and using x_m in the calculation is assuming g is the maximumum acceleration and that it occurs at that maximum permitted amplitude.

Ha!
And now back to stressing over everything else I need to know by Thursday morning...

The problem was posted on WebAssign, I just now checked back in the book (Halliday, Fundementals of Physics, 7e p. 405 #13) to see if any pertinent information was left out. The textbook's question in its entirety reads:
"A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to 1.00 [itex]\mu m[/tex]. (a) At what frequency is the magnitude <i>a</i> of the diaphragm's acceleration equal to <i>g</i>? (b) For greater frequencies, is <i>a</i> greater than or less than <i>g</i>?"[/itex]
 
Last edited:
mbrmbrg said:
Ha!
And now back to stressing over everything else I need to know by Thursday morning...

The problem was posted on WebAssign, I just now checked back in the book (Halliday, Fundementals of Physics, 7e p. 405 #13) to see if any pertinent information was left out. The textbook's question in its entirety reads:
"A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to 1.00 [itex]\mu m[/tex]. (a) At what frequency is the magnitude <i>a</i> of the diaphragm's acceleration equal to <i>g</i>? (b) For greater frequencies, is <i>a</i> greater than or less than <i>g</i>?"[/itex]
[itex] <br /> It should say <b>maximum magnitude</b> and <b>at the maximum possible amplitude</b>. As you can see from that cosine function, the magnitude is constantly changing for any given amplitude and frequency. At that maximum amplitude, the maximum acceleration at higher frequencies will be greater, but a <b>lower intensity</b> higher frequency sound might never have a = g.[/itex]
 

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