Amplitude and Pressure Wave: Pa Homework

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In summary, during a normal conversation, the amplitude of a pressure wave is 0.023 Pa. Using this information, we can calculate the force on the eardrum, which is 0.01288 N. The mechanical advantage of the three bones in the middle ear is 1.5, meaning the force exerted on the oval window by the bones is also 0.01288 N. Knowing the area of the oval window is 0.026 cm^2, we can determine that the pressure increase transmitted to the liquid in the cochlea is 0.496 Pa. This can be calculated using the equation dB=20log(P/P). To convert cm^2 to m^2, we must multiply by
  • #1
Webassign night again!

Homework Statement


During normal conversation, the amplitude of a pressure wave is 0.023 Pa.
(a) If the area of the eardrum is 0.56 cm2, what is the force on the eardrum?
N

(b) The mechanical advantage of the three bones in the middle ear is 1.5. If the force in part a is transmitted undiminished to the bones, what force do the bones exert on the oval window, the membrane to which the third bone is attached?
N

(c) The area of the oval window is 0.026 cm2. What is the pressure increase transmitted to the liquid in the cochlea?
Pa



Homework Equations


I'd assume I'd be using dB=20log(P/P) But I'm not sure how to apply that to any of these equations.


The Attempt at a Solution


Can't even figure out where to start. I'm sure with some starting advice, or at least how to solve the first part I could figure it out.
 
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  • #2
Maybe you could start with F = pressure * area
 
  • #3
well, to convert cm^2 to m^2 do I have to do anything special besides just dividing by 100?
 
  • #4
bnwchbammer said:
well, to convert cm^2 to m^2 do I have to do anything special besides just dividing by 100?

You have to do more than divide by 100. If it was converting cm --> m dividing by 100 would be ok.

Think of it this way in terms of multiplication- You want to cancel the units until you have m^2.

So-

1cm^2 *(1m^2)/(100^2 cm^2) = .0001 m^2.

So you know there are 100 cm in a meter. So if everything is squared, that must mean that there are 100^2 cm^2 in a m^2. So as you see above, when looking at the units, the cm^2's cancel out and you are left with m^2.
 
  • #5
Should'a been able to figure that out, but got it now. 100% on webassign, all right. Thanks for helpin me out guys.
 

1. What is amplitude in a pressure wave?

Amplitude is the maximum displacement or distance that a particle in a medium moves from its rest position when a pressure wave passes through it. In other words, it measures the strength or intensity of the wave.

2. How is amplitude related to pressure in a wave?

The amplitude of a pressure wave is directly proportional to the pressure of the wave. This means that as the amplitude increases, the pressure also increases, and vice versa.

3. What factors affect the amplitude of a pressure wave?

The amplitude of a pressure wave can be affected by various factors, such as the energy or source of the wave, the distance the wave travels, and the medium through which the wave travels. In general, the amplitude decreases as the wave travels further or encounters a medium with greater resistance.

4. How is amplitude measured in a pressure wave?

Amplitude is typically measured in units of pressure, such as Pascal (Pa) or Newtons per square meter (N/m2). It can also be measured in units of displacement, such as meters (m) or millimeters (mm).

5. What is the relationship between amplitude and frequency in a pressure wave?

The frequency of a pressure wave refers to the number of waves that pass through a point in a given time. The amplitude and frequency of a pressure wave are inversely proportional, meaning that as the frequency increases, the amplitude decreases and vice versa. This relationship is described by the wave equation: c = fλ, where c is the speed of the wave, f is the frequency, and λ is the wavelength.

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