A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional

Question

A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately what? Group of answer choices

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Kiệt Gia 4 years 2021-08-10T17:43:51+00:00 1 Answers 141 views 0

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  1. Sapo1
    0
    2021-08-10T17:45:32+00:00
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    Answer:

    0.05cos10t

    Explanation:

    X(t) = Acos(wt+φ)

    The oscillation angular frequency can be calculated using below formula

    w = √(k/M)

    Where K is the spring constant

    But we were given body mass of 5.0 kg

    We know acceleration due to gravity as 9.8m)s^2

    The lenghth of spring which stretches =10 cm

    Then we can calculate the value of K

    k = (5.0kg*9.8 m/s^2)/0.10 m

    K= 490 N/m

    Then if we substitute these values into the formula above we have

    w = √(k/M)

    w = √(490/5)

    = 9.90 rad/s=10rads/s(approximately)

    Its position as a function of time can be calculated using the below expresion

    X(t) = Acos(wt+φ)

    We were given amplitude of 5 cm , if we convert to metre = 0.05m

    w=10rads/s

    Then if we substitute we have

    X(t)=0.05cos(10×t)

    X(t)= 0.05cos10t

    Therefore,Its position as a function of time=

    X(t)= 0.05cos10t

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