Help with physics question about oscillation
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Help with physics question about oscillation

[From: ] [author: ] [Date: 11-06-28] [Hit: ]
e, we are going to neglect air friction, spring constant details, weight of spring, etc.).......
A spring with a mass on the end of it hangs in equilibrium a distance of 0.42 m above the floor. The mass is pulled down a distance of 0.06 m below the original position, released, and allowed to oscillate. How high above the floor is the mass at the highest point in its oscillation?

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Well, judging by how little information you are given, I am guessing this question is just testing your understanding of energy conservation (i.e, we are going to neglect air friction, spring constant details, weight of spring, etc.).

So if at rest the mass hangs 0.42 m above the floor, and you pull it 0.06 m below this position, then you are putting enough energy (into the stretched spring) so that upon release the weight will compress the spring and equal distance (i.e, 0.06 m) above the rest height. So at the highest point the weight will be 0.42 m + 0.06 m = 0.48 m above the ground.

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0,48 m above the floor.

You put enough elastic potential energy into the the spring to stretch it 0.06 m down to 0.36 m above the floor,. when you let it go the spring pulls the mass back up. When it gets back to 0.42 m there is no more elastic potential energy in the spring. It is all converted to kinetic energy. The velocity of the body is at maximum and it is moving up. It compresses the spring until all of the kinetic energy is converted back to elastic potential energy. It is assumed that it takes the same amount of energy to compress the spring 0.06 m as it does to expand it. So now the mass is at 0.48 m.
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