The displacement of the air molecules in a sound wave is modeled with the wave function s(x, t) = 3.00 nm cos(50.00 m−1x − 1.71 ✕ 104 s−1t).

Question

The displacement of the air molecules in a sound wave is modeled with the wave function s(x, t) = 3.00 nm cos(50.00 m−1x − 1.71 ✕ 104 s−1t). (a) What is the wave speed (in m/s) of the sound wave? 342 Correct: Your answer is correct. m/s (b) What is the maximum speed (in m/s) of the air molecules as they oscillate in simple harmonic motion? m/s (c) What is the magnitude of the maximum acceleration (in m/s2) of the air molecules as they oscillate in simple harmonic motion?

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Orla Orla 3 years 2021-09-02T18:06:39+00:00 1 Answers 56 views 0

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    2021-09-02T18:08:07+00:00

    Answer:

    a) 342 m/s

    b) 51*10^-6 m/s

    c) 0.87m/s^2

    Explanation:

    The following function describes the displacement of the molecules in a sound wave:

    s(x,t)=3.00nm\ cos(50.00\ m^{-1}x-1.71*10^4s^{-1}t)  (1)

    The general form of a function that describes the same situation is:

    s(x,t)=Acos(kx-\omega t)   (2)

    By comparing equations (1) and (2) you have:

    k: wave number = 50.00 m^-1

    w: angular frequency = 1.71*10^4 s^-1

    A: amplitude of the oscillation = 3.00nm

    a) The speed of the sound is obtained by using the formula:

    v=\frac{\omega}{k}=\frac{1.71*10^4s^-1}{50.00m^{-1}}=342\frac{m}{s}

    b) The maximum speed of the molecules is the maximum value of the derivative of s(x,t), in time. Then, you first obtain the derivative:

    \frac{ds}{st}=-\omega A sin(kx-\omega t)

    The max value is:

    v_{max}=\omega A

    v_{max}=(1.71*10^4s^-1)(3.00nm)=51300\frac{nm}{s}=51\frac{\mu m}{s} = 51*10^-6 m/s

    c) The acceleration is the max value of the derivative of the speed, that is, the second derivative of the displacement s(x,t):

    a=\frac{dv}{dt}=\frac{d^2s}{dt^2}=-\omega^2A cos(kx-\omega t)\\\\a_{max}=\omega^2 A

    Then, the maximum acceleration is:

    a_{max}=(1.71*10^4s^{-1})^2(3.00nm)=0.87\frac{m}{s^2}

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