Physics(Please respond)

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A 0.33-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to 0.26 m (relative to its unstrained length), the speed of the sphere decreases from 5.3 to 4.2 m/s. What is the spring constant of the spring?


I know that the formula is F=-kx
but i am not sure what to plug in since there are two sets of numbers plus the mass.

  • Physics(Please respond) -

    x1=A•sin(ω•t1),
    sin(ω•t1)=x1/A,
    cos(ω•t1)=sqrt(1-sin²(ω•t1)=
    =sqrt (1-(x1/A)²),
    v1=A•ω•cos(ω•t1)=
    =A •ω•sqrt (1-(x1/A)²).
    (v1/A•ω)²= 1-(x1/A)²,
    v1²/ω² =A² - x1²,
    A² = x1² + v1²/ω²,
    Similar to this
    A² = x2² + v2²/ω²,
    Therefore,
    x1² + v1²/ω²= x2² + v2²/ω²,
    x1² •ω² + v1²= x2²• ω² + v2²,
    v1²- v2² = (x2²-x1² ) •ω²
    ω² = (v1²- v2²)/( x2²-x1²),
    ω² =k/m,
    k =m•(v1²- v2²)/( x2²-x1²)=
    0.33•(28.09—17.64)/(6.76-1.44) •10^-2=
    = 0.33•10.45•100/5.32=64.82 N/m.

  • Physics(Please respond) -

    thank you!

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