Help determining the sound wave function

AI Thread Summary
The discussion focuses on solving a sound wave problem involving pressure and displacement functions. The participants calculate parameters such as displacement amplitude (s_m), wave number (k), and angular frequency (ω) for two scenarios with different air densities and sound speeds. Initial calculations suggested values for s_m and k, but participants identified and corrected typographical errors in their frequency and amplitude calculations. The final values for the first case were confirmed as s_m = 6.14 x 10^-9 m and for the second case as s_m = 5.89 x 10^-9 m. The conversation emphasizes the importance of accuracy in mathematical derivations related to sound waves.
jwxie
Messages
278
Reaction score
0

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 \bigtriangleup p_{s} = 4.0 mPa. If the displacement function of the wave is \[s(x,t) = s_{m} cos(kx-wt)\], what are
(a) \[s_{m}\]
(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) \[s_{m}\]
(e) k, and
(f) w


Homework Equations



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

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

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 \omega should be \[2\pi f\] which gives \[100\pi\]

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

and we then find k using this equation \[\bigtriangleup p_{max} = v^{2}\rho \kappa s_{max}\]<br /> 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 \[ s_{m}\] 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.
 
Last edited by a moderator:
Physics news on Phys.org
jwxie said:

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


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

ehild
 
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?
 
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
 
right,... i have a typo in calculation
thanks!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top