A spring-mass system has a spring constant of 3 N/m. A mass of 1 kg is attached to the spring, and the motion takes place in a viscous fluid

Question

A spring-mass system has a spring constant of 3 N/m. A mass of 1 kg is attached to the spring, and the motion takes place in a viscous fluid that offers a resistance numerically equal to two times the magnitude of the instantaneous velocity. There is zero external force. If the mass is pulled so that the spring is stretched 0.4 m and released (with initial velocity of zero), find the position, u, of the mass at any time t. What is the position, u, as t→+[infinity]

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Thanh Thu 3 years 2021-08-06T21:47:33+00:00 1 Answers 58 views 0

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  1. Delwyn
    0
    2021-08-06T21:48:51+00:00
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    Answer:

    Explanation:

    Given that,

    Spring constant k = 3N/m

    Mass attached m=1kg

    The motion takes place in a viscous fluid that offers a resistance numerically equal to two times the magnitude of the instantaneous velocity. This implies that the damping coefficient is ζ =2

    There is zero external force

    I.e F=0

    Extension e = 0.4m

    Generally, the equation of a mass spring system is give as

    mu” + ζ u’ + ku = F(t)

    Then, inserting the given datas

    u” + 2u’ + 3u = 0

    Solving this differential equation using D operator

    Then, the characteristics equation is

    D² + 2D + 3 =0

    Using formula method

    D = (—b ± √( b² —4ac) ) /2a

    a =1, b = 2 and c =3

    D = (—2 ± √(2²—4×1×3)) / (2×1)

    D = (—2 ± √(4-12) ) /2

    D = (—2 ± √-8 )/2

    D = (—2 ± 2√2 •i)/2

    D = —1 ± √2 •i

    Then, the particular solution

    is

    Up=Aexp(-t)Cos√2t+Bexp(-t)Sin√2t

    No homogeneous solution(Uh) since, F(t) = 0

    U(t) = Up + Uh

    Therefore,

    U(t)= Aexp(-t)Cos√2t+Bexp(-t)Sin√2t

    Initial condition given

    The initial velocity is 0, i.e U'(0)=0

    So, let find U’

    U’=-Aexp(-t)Cos√2t — A√2 exp(-t)Sin√2t —Bexp(-t)Sin√2t + B√2exp(-t)Cos√2t

    So,

    U'(0)=-Aexp(0)Cos√2•0 — A√2 exp(0)Sin√2•0 —Bexp(0)Sin√2•0+ B√2exp(0)Cos√2•0

    U'(0)=-Aexp(0)Cos√2•0 — A√2 exp(0)Sin√2•0 —Bexp(0)Sin√2•0+ B√2exp(0)Cos√2•0

    U'(0)=-A+B

    0 = – A+B

    A=B

    Also, second condition at t=0, u=4

    U(t)= Aexp(-t)Cos√2t+Bexp(-t)Sin√2t

    U(0) = Aexp(0)Cos√2•0+Bexp(0)Sin√2•0

    U(0) = A

    4 = A

    Since, A=B

    Then, B=4

    So, the general solution becomes

    U(t)= 4exp(-t)Cos√2t+4exp(-t)Sin√2t

    So, this is the position at any time

    b. Then, value of U(t) at t→+∞

    U(t)= 4exp(-t)Cos√2t+4exp(-t)Sin√2t

    U(∞) = 4exp(-∞)Cos∞ + 4exp(-∞)Sin∞

    Since exp(-∞) =0

    And 1≤CosX≤1, 1≤SinX≤1

    U(∞) = 4exp(-∞)Cos∞ + 4exp(-∞)Sin∞

    U(∞) = 0 + 0

    U(∞) = 0

    At infinity, the position is zero.

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