In a horizontal block-spring system, the block mass is m=0.5kg and the spring constant is k =2N/m. At t=?

In a horizontal block-spring system, the block mass is m=0.5kg and the spring constant is k =2N/m. At t=0.1s the block velocity is 0.263m/s and the acceleration is 0.216m/s2. Assuming that the system is undergoing SHM, write the expression for the evolution of block displacement from equilibrium point of spring in time

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  • Whome
    Lv 7
    6 years ago
    Best Answer

    General equation for SHM system

    x(t) = Asin(ωt + φ)

    x'(t) = Aωcos(ωt + φ)

    x"(t) = -Aω²sin(ωt + φ)

    ω = √(k/m)

    ω = √(2/0.5)

    ω = 2

    x'(0.1) = 0.263 = 2Acos(2(0.1) + φ)

    0.1315 = Acos(0.2 + φ)

    A = 0.1315 / cos(0.2 + φ)

    x"(t) = 0.216 = -2²Asin(2(0.1) + φ)

    0.216 = -4(0.1315 / cos(0.2 + φ))sin(0.2 + φ)

    0.216 = -0.526 tan(0.2 + φ)

    - 0.41064 = tan(0.2 + φ)

    -0.38965 = 0.2 + φ

    φ = -0.58965

    0.263 = 2Acos(0.2 - 0.58965)

    0.263 = 2A(0.9250)

    A = 0.142

    x(t) = 0.142sin(2t + .590)

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