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7 October, 15:09

The upper end of a 3.80-m-long steel wire is fastened to the ceiling, and a 54.0-kg object is suspended from the lower end of the wire. you observe that it takes a transverse pulse 0.0492 s to travel from the bottom to the top of the wire. what is the mass of the wire?

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  1. 7 October, 15:58
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    Given that the lenth write steel l = 3.80m fastened celling mass m = 54.0kg

    v = s/t

    v = 3.8/0.0492

    v = 77.23 m/s

    now, the formula for speed in a wire is

    v = (T/ÎĽ) ^1/2

    where T is the tension in the string and ÎĽ is the mass pee unit length.

    v = (Tl/m) ^1/2 ... (replace ÎĽ with m/l)

    now T = 550N

    v = [ 550 (3.8) / m ]^1/2

    squaring both sides

    v^2 = 550 (3.8) / m

    v^2 = 2090/m

    putting the value of v calculated above and solving

    m = 0.35 kg
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