Nick asked in Science & MathematicsPhysics · 4 years ago

physics oscillation help please?

A 0.36-kg mass is attached to a spring with spring constant 7.5 N/m and let fall. To the nearest hundredth of a meter what is the point where it 'stops'?


b) what is the amplitude of the resulting motion?

1 Answer

  • NCS
    Lv 7
    4 years ago
    Favorite Answer

    GPE becomes SPE after the mass falls "x."

    mgx = ½kx²

    x = 2mg / k = 2 * 0.36kg * 9.8m/s² / 7.5N/m = 0.94 m

    b) The amplitude is half that, or A = 0.47 m

    Bonus: mass weighs mg = 3.53 N

    and at full extension, the spring force is

    kx = 7.5N/m * 0.94m = 7.05 N = 2mg

    so the spring force oscillates between 0mg and 2mg and is equal to mg at the equilibrium position (which is what makes it the equilibrium position).

    Hope this helps!

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