# Sound wave propagating from water through steel

• Inertigratus
In summary, the conversation discussed the calculation of the amplitude of vibrations caused by a sound wave with a frequency of 2 MHz propagating through water and hitting a steel hull, using equations for intensity and acoustic impedance. The conversation also touched on the concept of root mean square and its application in calculating average power for AC voltage.
Inertigratus

## 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:
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!

Last edited:
"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

Ohh, I think I understand... thanks a lot!

## 1. How does sound travel through water and steel?

Sound travels through water and steel in the form of longitudinal waves. These waves cause the particles in the medium to vibrate in the same direction as the wave is propagating.

## 2. What factors affect the speed of sound in water and steel?

The speed of sound in water and steel is affected by the density and elasticity of the medium. In general, the denser and more elastic the medium, the faster sound will travel through it.

## 3. Can sound travel through water and steel at the same speed?

No, sound travels at different speeds in water and steel. The speed of sound in water is approximately 4 times faster than in air, while the speed of sound in steel is about 15 times faster than in air.

## 4. How does the temperature of water and steel affect the speed of sound?

The speed of sound in water and steel is directly proportional to the temperature of the medium. As temperature increases, the particles in the medium vibrate more vigorously, allowing sound to travel faster.

## 5. Are there any other mediums that sound can travel through besides water and steel?

Yes, sound can travel through a variety of mediums including air, liquids, and solids. The speed of sound will vary depending on the properties of the medium it is traveling through.

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