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Displacement amplitude

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data

    A travelling 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.

    2. Relevant equations

    Pressure amplitude is related to displacement amplitude by

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

    3. 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?
  2. jcsd
  3. May 8, 2010 #2


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    Homework Helper

    ∆P is sinusoidal, what is the maximum value that sine of anything can take?
  4. May 8, 2010 #3
    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??
  5. May 8, 2010 #4


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    Homework Helper


    If the maximum of sine is 1, then shouldn't sin(15.5x − 5.32 × 103t ) =1?
  6. May 9, 2010 #5
    Thank you so much! :redface:
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