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March 29, 2017

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A solid copper object hangs at the bottom of a
steel wire of negligible mass. The top end of the wire is
fixed. When the wire is struck, it emits sound with a fundamental
frequency of 300 Hz. The copper object is then submerged
in water so that half its volume is below the water
line. Determine the new fundamental frequency.
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  • physics - ,

    It is the question if you say it is.

    Partially submerging the copper object will reduce the tension in the steel wire. There will be a buoyancy force equal to half the volume of the copper object multiplied by the density of water. Let V be the volume of the copper and g be the acceleration of gravity. The specific gravity of copper is 8.92, and that of water is 1.00.

    Copper object weight = 8.92 V*g
    Buoyancy force = 1.00*(V/2)*g = 0.50 V*g
    Effective weight = 8.42 V*g
    The tension T gets multiplied by a factor 8.42/8.92 = 0.9439.

    The speed of waves in the wire AND the fundamental frequency get multiplied by the square root of this factor, or 0.9716.

    The new fundamental frequency becomes
    300*0.9716 = 291.5 Hz
    That is about half of a half-tone on the musical scale.

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