A transverse, wave travelling on a chord is represented by D=0.22sin (5.6x+34t) where D and x are inmeters and t is in seconds. For this wav

Question

A transverse, wave travelling on a chord is represented by D=0.22sin (5.6x+34t) where D and x are inmeters and t is in seconds. For this wave, determine; a) wavelength b) frequency c) velocity (both magnitude and direction) d) amplitude e) maximum and minimum speed of particles in the chord.

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Dulcie 3 years 2021-08-27T19:46:24+00:00 1 Answers 35 views 0

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  1. King
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    2021-08-27T19:47:44+00:00
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    Answer:

    a) λ = 1.12 m

    b) f = 5.41 Hz

    c) v = 154.54 m/s

    d) A = 0.22m

    e)

    v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}\\\\

    Explanation:

    You have the following equation for a wave traveling on a cord:

    D=0.22sin(5.6x+34t)     (1)

    The general expression for a wave is given by:

    D=Asin(kx-\omega t)    (2)

    By comparing the equation (1) and (2) you have:

    A: amplitude of the wave = 0.22m

    k: wave number = 5.6 m^-1

    w: angular velocity = 34 rad/s

    a) The wavelength is given by substitution in the following expression:

    \lambda=\frac{2\pi}{k}=\frac{2\pi}{5.6m^{-1}}=1.12m

    b) The frequency is:

    f=\frac{\omega}{2\pi}=\frac{34s^{-1}}{2\pi}=5.41Hz

    c) The velocity of the wave is:

    v=\frac{\omega}{k}=\frac{34s^{-1}}{0.22m^{-1}}=154.54\frac{m}{s}

    d) The amplitude is 0.22m

    e) To calculate the maximum and minimum speed of the particles you obtain the derivative of  the equation of the wave, in time:

    v_D=\frac{dD}{dt}=(0.22)(34)cos(5.6x+34t)\\\\v_D=7.48cos(5.6x+34t)

    cos function has a minimum value -1 and maximum +1. Then, you obtain for maximum and minimum velocity:

    v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}\\\\

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