What are the concepts needed to deal with this Oscillation problem?

Question

Oscillation problem: A vertical spiral spring has its lower end attached to a fixed horizontal surface and theupper end carries a platform. Assume that the mass of the spring and the platform is negligible. The force required to produce unit compression in the spring is k.
A mass of mass m has been placed on the platform and equilibrium position is reached distance d below before placing the mass.
The mass is then pushed down through a short distance x, and released, and vertical simple harmonic motions take place.

If m=6.0kg and k=1000Nm^-1, calculate the maximum amplitude of the oscillation if the mass is just to remain in contact with the platform at all times. Explain carefully how you obtain your answer.

I don't understand the topic oscillations at all, please try to explain to me. Please try to be more patient and explain the concepts, no answers in numerical form are needed but they help to explain.

Answer 1

For oscillations:

omega^2 x amplitude = acceleration

omega = 2 pi / time period of oscillation

For spring system, period = 2 pi x root (m / k)

omega = root (k / m)

Acceleration = (k / m)^2 x amplitude
= k x amplitude / m

Acceleration should not exceed gravity for contact to be maintained:

k x / m = g
x = mg / k

Answer 2

The "concepts" for your problem revolve around one fundamental law of physics: conservation of energy (COE). COE says we can neither create or destroy energy. In fact, the only thing we can do is convert it from one form to another.

For the compressed spring problem, when you smoosh it down a distance d (the compression) you build up potential energy (PE(d)) equal to the amount of work = W(d) = F(d)d = kd^2 = PE(d) you put into that compression process. And the F(d) is that kd you alluded to; where k is the compression coefficient and d = the compression due to the weight of the mass (w = mg).

The first bit of compression (d) came from the weight of the mass (m = 6 kg). The last (x) came from pushing down on the already compressed spring. So the work for this last bit of compression is W(x) = kx^2 = F(x) x; again we convert that work into more potential energy (PE(x) = kx^2).

Now we release the further compressed spring...what happens? The potential energy from that extra compression we gave the spring is converted into kinetic energy (KE(x) = PE(x)). So we have KE(x) = 1/2 mv^2 = kx^2 = PE(x) and as the spring decompresses, what happens to KE(x), it goes to zero and, once again, become potential energy when v = 0 At that point, PE(z) = kz^2 = PE(d - z) = k(d - z)^2; where z = the rebound distance (maximum amplitude) of the spring after it is released. z is the value you are looking for.

The two potentials are equal because they result from the same spring decompressed the same distance (z). But look at this, that KE(x) = PE(x) = 1/2 mv^2 = kx^2; so that PE(d - z) = KE(x), from the conservation of energy, and then PE(d - z) = k(d - z)^2 = KE(x) = kx^2 and we have (d - z) = x; so that z = d - x. Your spring will decompress until that extra KE(x) becomes potential energy at d - x from the point of full compression (d + x).

Because d - x < d, the spring would rebound to a point above the original PE(d) point. The difference in potentials PE(d) - PE(d - x) = KE(x) would be converted into kinetic energy as the spring (and weight) started to compress once more in the first of many oscillations.

Given the idealistic nature of your problem, the z you get would be the maxium amplitude and it would remain that value for each and every cycle. And given the same idealism, the oscillatiions would never cease unless you physically stop them somehow.

In real life, the amplitude of each oscillation (each cycle) would be less than the previous one because the conversion PE to KE would not be perfect...there would be heat losses due to metal fatigue and such; so not all that PE would go into KE. Eventually the heat losses would result in the oscillations stopping.