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Explain why, when the switch is closed and the magnet is oscillated vertically, the oscillations will dampen.

Here's my approach:

Assuming the bottom end of the magnet is the North pole, when this end moves downwards in the coil, the magnetic flux increases in the downwards direction. Thus, by Lenz's law, the resulting induced current must produce a magnetic field that is in the upwards direction inside the coil.

For this to happen, the current must be going counter-clockwise.

From here, I can't get to the conclusion that there will be an upwards force on the magnet when it moves downwards, and a downwards force on the magnet when it moves upwards.

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The upward movement would be produced by the spring stretched beyond the position of static equilibrium.

Lenz tells you that the direction of the induced current will always be such as to oppose the motion producing it.
As the magnet goes down the induced current will be such as to oppose the downward movement of the magnet.
That is, the magnet will feel an upward force upwards and have to do work to move down.
While all this is going on the induced current is passing through a circuit which has resistance and so heat is being generated in the circuit.
In terms of energy transfer within the system as the magnet moves down its gravitational potential energy is being converted to kinetic energy, energy stored in the spring and heat.
You have damped harmonic motion with the damping provided by the induced current.
What happens next depends on the degree of damping.
The magnet could eventually stop in the static equilibrium position for the system. The magnet could stop and then be pulled up by th extended spring and then the induced current would be such as to produce a downward force on the magnet whilst all the time generating heat in the circuit.
So you would get damped oscillatory motion until the final state of static equilibrium.

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