Finding Displacement Amplitude of a Pressure Wave

[roam]

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 ρ_air=1.21 kgm^–3.

Homework Equations

Pressure amplitude is related to displacement amplitude by

\Delta P_{max}= \rho v \omega s_{max}

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 \Delta P_{max}.

How can I find \Delta P_{max} 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?

 

[rock.freak667]

∆P is sinusoidal, what is the maximum value that sine of anything can take?

 

[roam]

∆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:

∆P_max = 20.0 sin(1)= 0.34

\Delta P_{max}= \rho v \omega s_{max}

0.34= 1.21 \times 343 \times 5320 S_{max}

S_{max} = \frac{0.34}{2207959.6}= 1.53 \times 10^{-7}

But this is not the correct answer, the correct answer must be 9.05 μm. What am I doing wrong here??

 

[rock.freak667]

I think the maximum value for sine is 1.

Yes.

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?

 

[roam]

Thank you so much!