Soundwave intensity due to altitude

[dnoi]

Homework Statement

Consider a flat Earth with an atmosphere that decreases in density as altitude increases such that p = p0*e$$^{-h/H}$$, where p0 is the density of air at zero altitude and H is a constant known as the "scale height." Assume the bulk modulus of air is constant.


a) Show that the intensity of a sound wave of constant wavelength will increase with altitude. (Hint: Assume the velocity changes due to a change in frequency only.)

b) Show that the intensity of a sound wave of constant frequency will decrease with altitude. (Hint: Assume the velocity changes due to a change in wavelength.)

c) Show that the ratio of the derivatives is given by

(dI/dh)$$\left|\lambda=constant$$
$$\overline{(dI/dh)\left|f=constant}$$ = -e$$^{h/H}$$





since i know that the density of air is decreasing with altitude, then that should mean that the density is decreasing as well. and if the density is decreasing, then the intensity of sound waves should decrease with altitude as well, right? since

I = 0.5*p*v*w^2*S^2

so is this just a matter of solving for p in the intensity equation and substituting it in for the equation given in the problem?

 

[AvroArrow]

or is there something more complicated that i'm missing? Homework Equations I = 0.5*p*v*w^2*S^2The Attempt at a Solution a) For this part, we can assume that the velocity of the sound wave changes due to a change in frequency only. Thus, we can write the intensity equation as: I = 0.5*p0*e^{-h/H}*v*w^2*S^2Taking the derivative with respect to h, we get: (dI/dh)\left|f=constant = -p0*v*w^2*S^2*e^{-h/H}/HSince the expression on the right is always negative, the intensity of the sound wave increases with altitude. b) In this case, we can assume that the velocity of the sound wave changes due to a change in wavelength only. Thus, we can write the intensity equation as: I = 0.5*p0*e^{-h/H}*v*w^2*S^2Taking the derivative with respect to h, we get: (dI/dh)\left|λ=constant = -p0*v*w^2*S^2*e^{-h/H}/HSince the expression on the right is always negative, the intensity of the sound wave decreases with altitude. c) The ratio of the derivatives for a sound wave of constant wavelength to a sound wave of constant frequency is given by: (dI/dh)\left|λ=constant \overline{(dI/dh)\left|f=constant} = -e^{h/H}

 

[Moetasim]

Yes, you are correct. The intensity of a sound wave is directly proportional to the density of the medium it is traveling through. In this case, the density of air is decreasing with altitude, which means the intensity of sound waves will also decrease. This can be seen by solving for p in the intensity equation and substituting it into the equation given in the problem.

For part a), assuming the velocity changes due to a change in frequency only, we can use the relationship between wavelength, frequency, and velocity (v = f*wavelength) to rewrite the intensity equation as I = 0.5*(p0*e^{-h/H})*(f*wavelength)^2*S^2. Since the frequency and wavelength are constant, the only variable that changes with altitude is the density (p). Thus, as altitude increases and p decreases, the intensity of the sound wave will increase.

For part b), assuming the velocity changes due to a change in wavelength, we can again use the relationship between wavelength, frequency, and velocity to rewrite the intensity equation as I = 0.5*(p0*e^{-h/H})*(f*wavelength)^2*S^2. In this case, the only variable that changes with altitude is the wavelength (wavelength = v/f). As altitude increases, the velocity decreases due to the decrease in density, which means the wavelength will also decrease. This leads to a decrease in intensity.

Finally, for part c), we can take the derivatives of the intensity equation with respect to altitude and use the chain rule to get (dI/dh)|λ=constant = -0.5*(p0/H)*e^{-h/H}*(f*wavelength)^2*S^2 and (dI/dh)|f=constant = -0.5*p0*e^{-h/H}*(f*wavelength)*(f*wavelength)*S^2. Dividing these two derivatives and simplifying, we get the desired ratio of -e^{h/H}. This shows that the rate of change of intensity with respect to altitude is greater for a constant wavelength than for a constant frequency, as shown in parts a) and b).