A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 10 sin(t/2)N (newtons) and moves in a medium that impa

Question

A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 10 sin(t/2)N (newtons) and moves in a medium that imparts a viscous force of 2 N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass.

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Eirian 4 years 2021-09-04T18:09:48+00:00 1 Answers 21 views 0

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  1. thaiduong
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    2021-09-04T18:10:48+00:00
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    Answer:

    Initial value problem is:

    u” + 10u’ + 98u = (2 sin(t/2) N for u(0) = 0

    u'(0) = 0.03m/s

    Explanation:

    The directions of Fd(t*) and U'(t*) are not specified in the question, so we’ll take Fd(t*) to be negative and U'(t*) to be positive. This is due to the fact that the damping factor acts in the direction opposite the direction of the motion of the mass.

    M = 5kg; L= 10cm or 0.1m;

    F(t) = 10 sin(t/2) N ; Fd(t*) = – 2N

    U'(t*) = 4cm/s or 0.04m/s

    u(0) = 0

    u'(0) = 3cm/s or 0.03m/s

    Now, we know that W = KL.

    Where K is the spring constant.

    And L is the length of extension.

    So, k = W/L

    W= mg = 5 x 9.81 = 49.05N

    So,k = 49.05/0.1 = 490.5kg/s^(2)

    Now from spring damping, we know that; Fd(t*) = – γu'(t*)

    Where,γ = damping coefficient

    So, γ = – Fd(t*)/u'(t*)

    So, γ = 2/0.04 = 50 Ns/m

    Therefore, the initial value problem which describes the motion of the mass is;

    5u” + 50u’ + 490u = (10 sin(t/2) N

    Divide each term by 5 to give;

    u” + 10u’ + 98u = (2 sin(t/2) N for u(0) = 0

    u'(0) = 0.03m/s

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