Pressure Variation in Sound Waves

[Cjho]

Homework Statement

Write an expression that describes the pressure variation as a function of x and t for the waves in air (0∘C) if the air molecules undergo a maximum displacement equal to the diameter of an oxygen molecule, about 3.0×10^−10m. Assume a sound-wave frequency of 55 Hz.
Express your answer in terms of the variables x, t, and appropriate constants using two significant figures.

Homework Equations

∆P = -∆Pmax cos(kx-wt)

∆Pmax= BAk = rho v^2 A K = 2pi rho v A f (where v is velocity)

V sound in air = 331.3 m/s + 0.6t (in celcius)

The Attempt at a Solution

Plugging into the formulas I get: v sound in air = 331.3m/s

so ∆Pmax = 2 pi (1.29 kg/m^3) 331.3m/s (3 x 10^-10m) 55 Hz = 4.43 x 10^-5

k = 2pif/v = 1.04
w = 2 pi f = 345.57

Final Answer : ∆P = - 4.4 x 10^-5 cos(1.0x - 350t)
I also tried with the ∆P =∆Pmax sin (kx - wt) version of the formula, and the website says both are wrong. I'm not really sure where I'm going wrong with this problem; thanks for any help!

 

[anathema]

Your answer is almost correct, but there are a few minor errors.

Firstly, the units for ∆Pmax should be Pa (Pascals), not just a number. So the correct expression would be ∆P = -4.4 x 10^-5 Pa cos(1.0x - 350t).

Secondly, you have used the wrong value for the velocity of sound in air. The correct formula is v = 331.3 m/s + 0.6t (in Celsius). Since the temperature is given as 0°C, the velocity should be 331.3 m/s. So the correct value for ∆Pmax would be 4.43 x 10^-5 Pa.

Lastly, you have used the wrong value for the angular frequency (w). The correct formula is w = 2 pi f = 2 pi (55 Hz) = 345.57. So the final answer would be ∆P = -4.4 x 10^-5 Pa cos(1.0x - 345.57t).