# 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'?

Update:

b) what is the amplitude of the resulting motion?

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- NCSLv 74 years agoFavorite 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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