Simple Harmonic Motion

[crazyasian]

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So i'm reading the Berkley review physics book and it says that potential energy for simple harmonic motion is the elastic potential energy because gravity does not play a role in horizontal movement. If the spring is vertical however, say it was hanging, would gravity play any role here, or would potential energy still be the elastic potential?

 

[plzNOCarribbean]

I think you would add the 2. KEi+PEi +PEi elastic = KE final + PE final

 

[PiBond]

So i'm reading the Berkley review physics book and it says that potential energy for simple harmonic motion is the elastic potential energy because gravity does not play a role in horizontal movement. If the spring is vertical however, say it was hanging, would gravity play any role here, or would potential energy still be the elastic potential?

it we were trying to solve for the elastic potential when a spring was vertical, we could use the mass of the attached object to find the distance (x) the spring was displaced using F = -kx. The force would just be the mass of the object times gravity. Then we could just plug into the formula PE = 1/2kx^2 and find our answer. The elastic potential energy of the spring will depend on the displacement and the spring constant only. When the spring's vertical, gravity is just supplying a force to displace the spring a distance (x) from equilibrium.

 

[sangria1986]

it we were trying to solve for the elastic potential when a spring was vertical, we could use the mass of the attached object to find the distance (x) the spring was displaced using F = -kx. The force would just be the mass of the object times gravity. Then we could just plug into the formula PE = 1/2kx^2 and find our answer. The elastic potential energy of the spring will depend on the displacement and the spring constant only. When the spring's vertical, gravity is just supplying a force to displace the spring a distance (x) from equilibrium.

This is true but Does not Answer his question. What if an object was dropped on the vertical spring from a height h above the spring? Then what? Indeed the op is correct. You would equate mgh (height from start above spring) = -mgx+1/2kx^2 which represents the final position of the object on the spring when it pushes down on the spring