An object with mass 4.0 is attached to a spring with spring stiffness constant 2

Question

An object with mass 4.0 is attached to a spring with spring stiffness constant 230 and is executing simple harmonic motion. When the object is 1.9

Explanation / Answer

for spring restorative force F = - k x m dv / dt = - k x dv = - k/m x dt dv = - k/m x {dx/v} v dv = - k/m dx integrating (v^2/2) = - k/m (x^2/2) + c (v^2) = - k/m (x^2) + c1 ---------------------- at x = a (amplitude) v = 0 (spring retreats after reaching amplitude ------------------------- 0 = - k/m (a^2) + c1 >>> c1 put back (v^2) = - k/m (x^2) + k/m (a^2) v^2 = k/m [a^2 - x^2] ----------------------------- given x = 0.02, v = 0.55 0.55^2 = k/m [a^2 - 0.02^2] a^2 = [(m/k) * 0.55^2] + 0.02^2 calculate (a) amplitude -------------------------------- v^2 = k/m [a^2 - x^2] velocity will be maximum when x = 0 ie at mean position by seeing v^2 formula you can know that (- x^2) term is trying to decrease the v^2 so if x=0 this term will not reduce v^2 hence vmax (vmax)^2 = k/m * a^2 just found value of a is known and others are known vmax = a v(k/m) Source(s): hope you will settle these questions

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