What Determines the Amplitude of a Vibrating Membrane?

[furtivefelon]

hi, this is the last question I'm having trouble with.. the question is:

Fine sand is sprinkled on the horizontal membrane, which vibrates in the vertical direction with frequency f = 500 Hz. The sand particulates bounce at the height of 3 mm over the level of the position of equilibrium of the membrane.
What is the amplitude of the membrane’s vibration?

I imagine that as each sand particulates hits the membrane, the membrane would vibrate for several seconds, and it would be much lower in amplitude than the height sand bounces.. However, I don't have a clue currently how to go about finding that specific amplitude.. Can someone give me a clue as where to begin?

Thank you very much

 

[Nate810]

The amplitude of the membrane's vibration can be calculated using the equation: A = (h/2)*(f/f_0)^2, where h is the height of the sand particulates bouncing over the level of the position of equilibrium, f is the frequency of the vibration, and f_0 is the resonant frequency of the membrane. Therefore, the amplitude of the membrane's vibration would be A = (3mm/2)*(500Hz/f_0)^2.

 

[quicknote]

To find the amplitude of the membrane's vibration, we need to first understand what amplitude means in this context. Amplitude refers to the maximum displacement of the membrane from its position of equilibrium, in this case, the level of the membrane when it is not vibrating.

In this scenario, the sand particulates are causing the membrane to vibrate at a frequency of 500 Hz. This means that the membrane is vibrating 500 times per second. Each time the membrane vibrates, the sand particulates bounce 3 mm above the level of the membrane's equilibrium position.

To find the amplitude, we can use the formula A = d/2, where A is the amplitude and d is the maximum displacement. In this case, d = 3 mm, so the amplitude would be A = 3 mm/2 = 1.5 mm. Therefore, the amplitude of the membrane's vibration is 1.5 mm.

I hope this helps!