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11 May, 03:15

At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 29.8981 cm on a spring with a spring constant of 19.1461 N/m. The mass of the bananas is 43.288 kg. What is the maximum speed of the bananas

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  1. 11 May, 03:33
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    Given that,

    The amplitude of oscillation is

    A = 29.8981cm = 0.298981m

    Spring constant k = 19.1461N/m

    Mass of banana hung m = 43.288kg

    Maximum speed v?

    The maximum speed can be determined by using the formula

    v = w•A

    Where

    w = angular frequency

    A = amplitude

    So we need to get the angular frequency and it can be determined by using the formula

    w = √ (k/m)

    w = √ (19.1461/43.288)

    w = √0.4423

    w = 0.6651rad/sec

    Then, maximum speed

    v = w•A

    v = 0.6651 * 0.298981

    v = 0.19884m/s

    v ≈ 0.2m/s

    The maximum speed of oscillation is 0.2m/s
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