Biophysics: Amplitude of the pressure wave.

[remakm]

Homework Statement

Say that the logarithmic intensity of a dolphin's sound is 50dB as measured as 1m away. What would be the ampltude of the pressure wave at 20m from the dolphin?

My Question/Problem: How are you supposed to factor in the 20m away?

Homework Equations

P_o= SQrt(2*Density*Velocity*Intensity)

Intensity = 10Log(I/(10^-16W/cm^2))

The Attempt at a Solution

Intensity:
50dB = 10Log(I/10^-16W/cm^2)
5.0dB=Log(I/10^-16W/cm^2)
10^5.0=(I/10^-16W/cm^2)
I=10^16/10^5 = 10^11W/cm^2

This is where I am stuck because I think this is where the 20m factors in, I just don't know how exactly. What I am thinking is before I solve for I, I divide 50dB/20m?? Is that correct?

Once I get the correct value for I, I just plug and chug into the Po equation. If I a completely wrong in my thinking, please guide me in the correct direction. Thank you for you help and time.

 

[collinsmark]

Hello remakm,

Welcome to Physics Forums!

Intensity:
50dB = 10Log(I/10^-16W/cm^2)
5.0dB=Log(I/10^-16W/cm^2)
10^5.0=(I/10^-16W/cm^2)
I=10^16/10^5 = 10^11W/cm^2

Double check that very last line in your calculation. I think you divided when you should have multiplied, or used the wrong sign on one of your exponents, or something like that.

This is where I am stuck because I think this is where the 20m factors in, I just don't know how exactly. What I am thinking is before I solve for I, I divide 50dB/20m?? Is that correct?

Once I get the correct value for I, I just plug and chug into the Po equation. If I a completely wrong in my thinking, please guide me in the correct direction. Thank you for you help and time.

From here there are actually a couple of ways you could proceed. But I'll try to stick with the basics to start.

Once you find I you'll know the power per unit area. What's the surface area of sphere of radius 1 meter? Given that, what's the total power?

After you've figured out the total power, ask yourself what is the surface area of a sphere of radius 20 m? So what's the power per unit area 20 m away? (Hint: remember the total power hasn't changed -- only the surface area of the sphere has changed.)

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Solve the problem in the above way first. But later, you might find it useful rework the problem a second time (or third time), keeping in mind some useful things about logarithms and decibels.

\log (x^2) = 2 \log x

And since I \propto P^2, [/tex] (for constant density, velocity, etc.) [Edit: and just for clarity, I'm using P [/tex] to represent pressure.]<br /> <br /> 10 \log \left( \frac{I}{I_o} \right) = 10 \log \left( \frac{P^2}{P_o^2} \right) = 20 \log \left( \frac{P}{P_o} \right)<br /> <br /> For a given total power, there is also a relationship involving <i>I</i> and the distance <i>r</i> away from the source (hint: think back to the radius of a sphere). Can you use a similar trick to the above which involves <i>r</i>? <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /> If you've bothered to follow through up to this point, you'll thank yourself. Finding this relationship (involving how different distances <i>r</i> give a corresponding change in dB), and then reworking the problem makes the problem almost trivial.