PHY young modulus+ wave一問 急!!

1. Two wire one of steel(YM:2.0X10^11 Pa) and one of copper(YM:1.3X10^11Pa) each of length 1.5m, diameter 2.5mm join end to end and form composite wire 3.0m what is total extension of composite wire under tension 100N?

ANS is 0.39 mm HOW?

2. if there is a light beam composed of 2 mpnpchromatic components a ,b is incident on an ordinary rectandular glass block. It eventually splits into 2 monochromatic beams with beam a having lower frequency than b 點解係e個ray correct? why a bend more?

圖片參考:http://imgcld.yimg.com/8/n/HA00213654/o/7012041100...

更新:

THX 但我唔係好明freq 細d n細d咩? f1/f2 proportion to n2/n1 ??

1 個解答

評分
  • 天同
    Lv 7
    9 年前
    最愛解答

    1. Young' Modulud E = stree/strain = (F/A)/(x/L)

    where F is the tension in the wire

    A is the cross-sectional area

    x is the extension

    L is the natural length of the wire

    Hence, x = F.L/(E.A)

    Consider the steel wire, cross-sectional area A = pi x (2.5/2 x 10^-3)^2 m^2 = 4.91 x 10^-6 m^2

    Extension x1 = 100 x 1.5/(2x10^11 x 4.91x10^-6) m

    Consider the copper wire, extension x2 = 100 x 1.5/(1.3x10^11 x 4.91x10^-6) m

    Hence, total extension of the composite wire = x1 + x2

    = (100 x 1.5/4.91x10^-6) x [1/(2x10^11) + 1/(1.3x10^11)] m

    = 3.9x10^-4 m = 0.39 mm

    2. It is given that beam a is of frequency lower than beam b. Thus the refractive index of beam a is smaller than that of beam b.

    Therefore, when the two beams incident onto the glass block, beam a (because of its small refractive index), bends less than beam b.

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