Question

Different mass crates are placed on top of springs of uncompressedlength and stiffness . The crates are released and the springscompress to a length before bringing the crates back up to theiroriginal positions.

Answer 1

yes it is right

Answer 2

The general solution for simple harmonic motion is:

X = A cos(wt + p)

In our case, the initial position is the highest so we can take p to be 0.

The mass gets back to its initial position after 2 pi radians, so the period P is given by:

wP = 2 pi or P = (2)(pi)/w

So the higher w the shorter the period. So we need to determine w in terms of k, L, and L0.

The page cited above notes that:
w = sqrt(k/m)

So we can determine w if we can determine the mass m.

Now we know from Hooke's law that the energy in a spring is given by:

U = (1/2)kx^2

(1/2)k(L0 - L)^2 = mg(L0 - L)

This lets you compute m from k, L0, and L:

m = (1/2)k(L0-L)/g

Answer 3

Nice little problem.

A mass on a spring is the classical example of simple harmonic motion:
http://en.wikipedia.org/wiki/Harmonic_os…

The general solution for simple harmonic motion is:

X = A cos(wt + p)

In our case, the initial position is the highest so we can take p to be 0.

The mass gets back to its initial position after 2 pi radians, so the period P is given by:

wP = 2 pi or P = (2)(pi)/w

So the higher w the shorter the period. So we need to determine w in terms of k, L, and L0.

The page cited above notes that:
w = sqrt(k/m)

So we can determine w if we can determine the mass m.

Now we know from Hooke's law that the energy in a spring is given by:

U = (1/2)kx^2

We also know that:
1. energy in this system is conserved

2. the three energies in this system are the potential energy due to gravity, the kinetic energy, and the spring energy.

3. At the initial state, the velocity of the mass is 0 so its kinetic energy is 0, the spring is uncompressed so its energy is 0, and the height is 0 = so the sum of all energies is 0.

4. At the fully compressed state, the velocity is still 0 and the mass has moved down by a distance (L0 - L) while the spring has compressed by a distance (L0 -L).

Thus the change in potential energy is matched by that in spring energy, or:

(1/2)k(L0 - L)^2 = mg(L0 - L)

This lets you compute m from k, L0, and L:

m = (1/2)k(L0-L)/g

With m and k for each case, you can compute w and sort the items as desired.