# simple pendulum question

Discussion in 'MCAT Study Question Q&A' started by zzzcor, Jul 19, 2011.

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1. ### zzzcor 2+ Year Member

89
0
Oct 23, 2009
let's say a mass is hanging down from the ceiling at certain distance from the ceiling and say it's equilibrium distance is x from the ceiling when mass is at the rest.

if k = 20N/m what is maximum distance that can be displaced from x when 250J is applied..

apparently I set 1/2 kx^2 = 250

and x = 5m
but I thought it should be less than 5m because as the mass goes up doesn't the the gravity and spring both acts on the mass?

So at equilibrium distance as it goes up, there is no force acting on it but as soon as it goes up beyond the equilibrium point, there is potential energy and elastic energy to count so that maximum distance should be less than that?

so 250 = mgh + 1/2kx^2 which in this case x = h since they are basically the same?

the answer says it's 5 m though from GS FL's. some questions from here are whacks.

please clarify.

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3. ### zwander 5+ Year Member

344
0
May 12, 2010
Château d'If
Yea this is kind of tricky. We I guess my explanation would be that before you add potential energy to the spring, it's at its equilibrium position where the force of gravity is equal to the restoring force on the spring (a=0). So if you are to add 250 Joules of energy into the system, a spring with that spring constant would have to extend 5m according to PE=1/2kx^2. By extending the spring, you've increased the spring restoring force F=kx for there would be a net upward force. They at the equilibrium position mg again eqauls kx, but then about the equilibrium position, you have mg>kx so the spring begins to accelerate in opposite direction. Then finally at the other equilibrium position x above the spring, PE is again max, v=0 and you have maximum acceleration in the opposite direction and the mass returns to its initial position and the cycle repeats.