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If the bell is still vibrating when you let air inside it, then the answer is yes. If the bell was damped just before the door is opened, then the answer is no. Sound is transmitted through compression / decompression waves (pressure waves) in a medium (e.g. air, water, wall). This necessitates contact of the vibrating source of sound with such a medium. ...

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Much of what you ask is covered in the question Demonstration that vibrating basic particles constitute non-vibrating individuals. The vibration is imperceptible to the eye so the object wouldn't look different. However you ask: Also, would this type of vibration cause any negative effects to the object? Yes, indeed it would. The vibration you ...

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The maximum acceleration experienced by your accelerometer (assuming that it does not automatically zero out the acceleration due to gravity) will be the same as the maximum acceleration experienced by any other object in the helicopter, as long as the helicopter attitude is maintained (no rotation). You would simply take the instantaneous acceleration ...

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The trouble is that your table, or whatever object it is, will act as a waveguide. That's because the sound waves will (partially) reflect of the wood/air surface then travel back into the table and interfere with other waves. The result is going to be hideously complicated to calculate. As Luboš says in a comment, if the thickness of the table is much less ...

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In today's news - researchers at MIT did just that using high-speed camera with frame rate between 2 kHz and 6 kHz. They used some advanced filtering to detect microscopic movement of objects, but for details we will have to wait until they publish their paper.

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I think you mean: for a given loudness, which frequencies involve greater physical movement, high frequencies or low? And that is simple - the lower the frequency the greater the amplitude of the movement. Here's a simple demonstration. Take the grille cloth off a speaker with a woofer. Play music through the speaker, something with sustained notes like ...

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No, the resonance frequancy is just dependent on your forcing frequency $\omega$ and the attenuation $c$. If you start with $$m\ddot{x}+c\dot{x}+k x = F_0 \cos{(\omega t)}$$ you will get something like $$|A| = \frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2 + 4r^2\omega^2}}$$ for your amplitude $A$, where $r=\frac{c}{2m}$. With $F_0 = \text{const.}$ just ...

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