Sound wave propagating from water through steel

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Homework Help Overview

The discussion revolves around a sound wave with a frequency of 2 MHz propagating from water into a steel ship hull. Participants explore the relationship between acoustic pressure, impedance, and intensity as the wave transitions between the two media, with specific values provided for acoustic impedance and wave velocities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relevance of equations related to intensity and amplitude, questioning whether intensity remains constant during the transition from water to steel. Some attempt to calculate transmitted intensity and amplitude based on given parameters, while others express uncertainty about the application of RMS pressure in their calculations.

Discussion Status

There is an ongoing exploration of the calculations involved, with some participants successfully arriving at the correct amplitude while others are still verifying their results. Clarifications regarding the use of RMS values and the implications of acoustic impedance are being discussed, indicating a productive exchange of ideas.

Contextual Notes

Participants are navigating through concepts of acoustic pressure and intensity, with some expressing confusion about the application of RMS values in their calculations. There is also mention of reflection and transmission at the boundary between water and steel, which adds complexity to the problem.

Inertigratus
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Homework Statement


A sound wave with frequency 2 MHz is propagating through water and hits a ship hull made out of steel. The effective value (RMS?) of the acoustic pressure in water is 1kPa = P0w.
The acoustic impedance Zw for water is 1.46*106 and for steel Zs it is 4.03*107. The wave propagation velocity is 1500 in water and 3800 in steel.
Find the amplitude (S0?) of the vibrations that the wave is causing through the steel hull.

Homework Equations


I = P02/2Z = (Z/2)*S022

The Attempt at a Solution


I have no idea how to do this. The only equation that seemed relevant to me is the one above. Perhaps the intensity is constant while the wave goes from water to steel?
If so, we could express the left part of the equation with the variables from when the wave is in the water and the right part with the variables from when it's in the steel hull.
I tried it though, and it's wrong :/

Ok, I think I figured it out... took some time. Had to calculate the transmitted intensity and then use the equation to the right to solve for the amplitude. Still a little off from the right answer, going to check the calculations again.

Well it's on the right magnitude atleast, I'm getting 3.8 * 10^-12 and it's supposed to be 5.4 * 10^-12. Any ideas?
 
Last edited:
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Do not forget that part of the sound wave is reflected from the water-steel boundary and the other part enters into the steel. You can calculate the reflectance and transmittance from the acoustic impedance.
http://www.ndt-ed.org/EducationResources/CommunityCollege/Ultrasonics/Physics/reflectiontransmission.htm

ehild
 
Thanks, but that's what I did :). The transmitted intensity is the incoming minus the reflected. The reflected is R times the incoming. Then I solved for the amplitude from the equation above... but it's a little bit off from the answer.
 
Have you taken into account that the rms pressure is given in water, and the amplitude is asked in steel? I=P2rms/Z.

ehild
 
Yes, I used that equation to find the intensity of the "incoming" wave, then calculated the transmitted intensity and then from that solved for the amplitude using the other equation.
 
I got 5.4 * 10^-12. What is your incoming intensity?

ehild
 
Nice! That's the right answer, how did you get it?
My incoming intensity is 3.4 * 10^-1.
Got it from I = (1000)^2 / 2*(1.46 * 10^6)
 
Inertigratus said:
Nice! That's the right answer, how did you get it?
My incoming intensity is 3.4 * 10^-1.
Got it from I = (1000)^2 / 2*(1.46 * 10^6)

You do not need to divide by 2 as the rms pressure is given.

ehild
 
Really? But that's the formula? How come I don't have to divide by 2?
Rms is pretty new to me, just recently heard about it.
I'm not questioning you, just want to know why that is.

Also, if you have the time. Could you shortly describe how I could calculate the thickness of a piezoelectric crystal when knowing the frequency, elasticity modulus and density?

Thanks!
 
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  • #10
"rms" means root mean square value of a periodic quantity, X(t), the square root of the time average of the square of X(t).
You have something, the pressure in this problem, that varies sinusoidally in time: P=P0sin(wt). The power at an instant is (P0sin(wt))2/Z but it changes very fast, and the average power counts really. It is the same with the AC current and voltage, your heater, hair drier and so on needs to provide some average power and they are specified with that. In electricity, the rms value of the AC voltage means that DC voltage which would provide the same power on a resistor as the AC voltage does. You know that the power in case of DC voltage V across a resistor is V2/R.
You get the time average of the power by integrating the instantaneous one for one period and then dividing with the length of the period. The time average of the sin2(t) function is 1/2. So the average power of the AC voltage which has the amplitude V0 is V02/(2R) This power is equivalent with that of a DC source with voltage Vrms:

Vrms2/R=V02/(2R), that is Vrms=V0/√2

As for your other question, concerning the piezoelectric crystal, I can not answer.

ehild
 
  • #11
Ohh, I think I understand... thanks a lot!
 

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