Help determining the sound wave function

[jwxie]

Homework Statement

[PLAIN]http://dl.dropbox.com/u/14655573/110218_203248.jpg

Part (1)
The figure shows the output from a pressure monitor mounted at a point along the path taken by a sound wave of a single frequency traveling at 343 m/s through air with a single frequency traveling at 343 m/s through air with a uniform density of 1.21 kg/m^3. The vertical axis scale is set by [itex]\bigtriangleup p_{s}[/itex] = 4.0 mPa. If the displacement function of the wave is [itex]\[s(x,t) = s_{m} cos(kx-wt)\][/itex], what are
(a) [itex]\[s_{m}\][/itex]
(b) k, and
(c) w

Part (2)
The air is then cooled so that its density is 1.35 kg/m^3 and the speed of a sound wave through it is 320 m/s. The sound source again emits the sound wave at the same frequency and the same pressure amplitude. What now are
(d) [itex]\[s_{m}\][/itex]
(e) k, and
(f) w

Homework Equations

The pressure wave is the derivative of the displacement wave, and the pressure amplitude [itex]\bigtriangleup p_{max}[/itex] is relates to the displacement amplitude [itex]\bigtriangleup s_{max}[/itex] by :

[itex]\[\bigtriangleup p_{max} = v^{2}\rho \kappa s_{max}\]
[/itex]
or we can reduce further
[itex]\bigtriangleup p_{max} = v\rho \omega s_{max}[/itex]

The Attempt at a Solution

First, I think the period is 2 ms according to the graph. So the frequency should be 1/0.002 or 500Hz, and [itex]\omega[/itex] should be [itex]\[2\pi f\][/itex] which gives [itex]\[100\pi\][/itex]

Then equate 0.008 Pa to find [itex]\[s_{m}\][/itex]
[itex]\[0.008 = v^{2}\rho \omega s_{m}\][/itex]
I think I get 6.14 x 10^-8 m

and we then find k using this equation [itex]\[\bigtriangleup p_{max} = v^{2}\rho \kappa s_{max}\]
[/itex] and I got 0.9888 m^-1

For part 2, since frequency and pressure amplitude remains the same, so omega will same the same. K, however, will change because wavelength changes with speed.
and doing similar calculations, for [itex] \[ s_{m}\][/itex] is 5.89 x 10^-9 and k is 9.825 m^-1.

Do they even make sense? I see that the latter case is a magnitude higher...

I don't have any solutions to this problem. If anyone can verify this with me I will really appreciate! Thanks.

 

[ehild]

and doing similar calculations, for [itex] \[ s_{m}\][/itex] is 5.89 x 10^-9 .

Check it. It should be 5.89 x 10^-8 .

ehild

 

[jwxie]

hi, thanks ehild.
i see i had a typo in the first post
omega should be 1000pi, not 100pi
using 100pi would give the 8th order, but 1000pi would give 9th order
what do you think?

 

[ehild]

Yes, omega is 1000 pi. With that, smax =6.14 x 10^-9 m in the first case, and 5.89 x 10^-9 m in the second case.

ehild

 

[jwxie]

right,... i have a typo in calculation
thanks!