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## Homework Statement

A perfectly elastic spring swings in a vertical plane as a simple pendulum with a mass m suspended at the bottom of the spring. The force constant for this spring is ##k## and the unstretched length is ##L##. The spring is carefully held in the horizontal position so that the spring is unstretched when released. The spring then swings through the vertical position. What is the increase in the length of the spring when the spring passes through the vertical position?

## Homework Equations

##E_i = E_f## (conservation of mechanical energy)

##\frac{mv^2}{r} = F_c##

## The Attempt at a Solution

It says that this is a simple pendulum, and doesn't specify any block width, so I'll assume that the mass is a point mass. I'll have ##\Delta x := x## to make it easier to type.

I know that mechanical energy is conserved so ##mg(L + x) = \frac{1}{2}mv^2 + \frac{1}{2}kx^2## at the lowest point.

I need to get rid of ##v## in my equation and the only other thing I can think of is centripetal force. I question if I can actually use this since I don't think this system moves with a constant distance to the pivot; I think I would need to use the radius of curvature for the parabola (?) that results. I'm not sure so I have just tried to use centripetal force directly: ##kx - mg = \frac{mv^2}{L+x}##. However, once I get this, combining everything gives me a very unwieldy quadratic that doesn't correspond to any of the given answer choices.