Finding Displacement Amplitude of a Pressure Wave

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roam
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Homework Statement



A traveling sound wave causes a variation in air pressure according to the equation:

∆P = 20.0 sin(15.5x − 5.32 × 103t )

where x is in metres, t in seconds and pressure is in pascals.

What is the amplitude of the displacement of the air particles caused by this pressure wave (i.e. the displacement amplitude)?

Take ρair=1.21 kgm–3.

Homework Equations



Pressure amplitude is related to displacement amplitude by

[tex]\Delta P_{max}= \rho v \omega s_{max}[/tex]

The Attempt at a Solution



I know that the angular frequency ω of the pressure wave is 5320.0 rads–1, and the velocity v of the pressure wave is 343.0 ms–1. But I can't use the formula above because I can't determine the value of [tex]\Delta P_{max}[/tex].

How can I find [tex]\Delta P_{max}[/tex] from

∆P = 20.0 sin(15.5x − 5.32 × 103t )

when I don't know the values of "x" and "t"? What values do I need to substitute there?
 
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rock.freak667 said:
∆P is sinusoidal, what is the maximum value that sine of anything can take?

I think the maximum value for sine is 1. So we must have:

(15.5x − 5.32 × 103t ) =1

Okay, then we get:

∆Pmax = 20.0 sin(1)= 0.34

[tex]\Delta P_{max}= \rho v \omega s_{max}[/tex]

[tex]0.34= 1.21 \times 343 \times 5320 S_{max}[/tex]

[tex]S_{max} = \frac{0.34}{2207959.6}= 1.53 \times 10^{-7}[/tex]

But this is not the correct answer, the correct answer must be 9.05 μm. What am I doing wrong here??
 
roam said:
I think the maximum value for sine is 1.

Yes.


roam said:
So we must have:

(15.5x − 5.32 × 103t ) =1


If the maximum of sine is 1, then shouldn't sin(15.5x − 5.32 × 103t ) =1?