The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in centimete

Question

The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string

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Philomena 4 years 2021-07-17T11:09:52+00:00 1 Answers 218 views 0

Answers ( )

  1. lethu
    0
    2021-07-17T11:11:49+00:00
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    Answer:

    given

    y=6.0sin(0.020px + 4.0pt)

    the general wave equation moving in the positive directionis

    y(x,t) = ymsin(kx -?t)

    a) the amplitude is

    ym = 6.0cm

    b)

    we have the angular wave number as

    k = 2p /?

    or

    ? = 2p / 0.020p

    =1.0*102cm

    c)

    the frequency is

    f = ?/2p

    = 4p/2p

    = 2.0 Hz

    d)

    the wave speed is

    v = f?

    = (100cm)(2.0Hz)

    = 2.0*102cm/s

    e)

    since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction

    f)

    the maximum transverse speed is

    umax =2pfym

    = 2p(2.0Hz)(6.0cm)

    = 75cm/s

    g)

    we have

    y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]

    = -2.0cm

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )