So step one is don't put number in equations until the end. And, if you do put numbers in: include units. Something like $(2)(3)^2$ is meaningless, and I've done hundreds of spring problems.

Step 2: choose a good coordinate system. The reasonable choices are:

1) $x=0$ is the top of the unloaded spring

2) $x=0$ is the equilibrium position.

I like (2), since the answer to (c) is $x=0$...but there is a reason to chose (1):

It's easy to write the potential energy as a function of $x$ (oh, and one must ask: is $x$ the right name for a vertical displacement? No, it's not. But I'm going to begrudging use that. $z$ is better, but this really is a 1-D problem, so we'll let it slide--actually, I have a reason).

There are two potentials, gravity

$$ V_g(x) = mgx $$

And a time dependent potential, that changes when the mass sticks to the spring, and that is:

$$ V_s(x) = \frac 1 2 k x^2 $$

The time coordinate can be define to that $t=0$ is when the mass hits the spring, or when it reaches equilibrium, or when the whole things starts. I like the penultimate, but will use the last.

So, at $t=0$:

$$ x(0) = 0.45\,{\rm m} \equiv x_0 $$

The total energy is (remember, the spring potential is "off"):

$$ E_0 = T(t=0) + V(t=0) = V_g(x_0) = mgx_0 $$

The time it takes to hit the spring, $t_1$, is a basic formula from:

$$ x_0 = \frac 1 2 g t_1^2 $$

so:

$$ t_1 = \sqrt{\frac{2x_0}{g}}$$

with (energy conservation):

$$ E_1 = T_1 + V_1 = E_0 $$

$$ \frac 1 2 mv^2_1 + 0 = mgx_0 $$

with

$$ v_1 = gt_1 = \sqrt{2gx_0} $$

Next, is calculate the equilibrium position, $x'$, which is the solution to:

$$ \frac{dU(x)}{dx} = 0 $$

$$ mg + kx' = 0 $$

$$ x' = -\frac{mg} k $$

Note that the energy of the compresses spring at equilibrium is:

$$ V_k(x') = \frac 1 2 k \frac{(mg)^2}{k^2} = \frac 1 2 \frac{(mg)^2}{k}$$

While the gravitational potential different from $x=0$ is

$$ V_g(x') = mgx' = \frac{(mg)^2}{k} = 2V_k(x') $$

That $V_g/V_k = 2$ in equilibrium comes up a lot in physics problem (remember it). If you consider the harmonic oscillator your bound system and "gravity" to be an external perturbing field, you'll see this in all kinds of systems, such as a polarizable atom in an external uniform field.

From here: write the equations of motion:

$$ ma - F = 0$$ $$ m\ddot x - kx - mg = 0 $$

and transform to $y = x-x' = x +\frac{mg} k $:

$$ m\ddot y - k(xy-\frac{mg} k) + - mg = 0 $$ $$ m\ddot y - kxy+\frac{kmg} k + - mg = 0 $$ $$ m\ddot y - ky = 0 $$

Notice how gravity has completely dropped out of the problem. This is unique (afaik) to the SHO.

So now solve that equation with the correct initial conditions: You know $y(t_1) = +|x'|$ and $\dot y(t_1) = v_1$.

So to reiterate: do not use real numbers until the end. Use physical quantities as variables, with clear and obvious names. Mark significant events with subscripts. At each step, think about "what conserved quantity do I know at that this step?", and "how does it get me to the next step?".

So step one is don't put number in equations until the end. And, if you do put numbers in: include units. Something like $(2)(3)^2$ is meaningless, and I've done hundreds of spring problems.

Step 2: choose a good coordinate system. The reasonable choices are:

1) $x=0$ is the top of the unloaded spring

2) $x=0$ is the equilibrium position.

I like (2), since the answer to (c) is $x=0$...but there is a reason to chose (1):

It's easy to write the potential energy as a function of $x$ (oh, and one must ask: is $x$ the right name for a vertical displacement? No, it's not. But I'm going to begrudging use that. $z$ is better, but this really is a 1-D problem, so we'll let it slide--actually, I have a reason).

There are two potentials, gravity

$$ V_g(x) = mgx $$

And a time dependent potential, that changes when the mass sticks to the spring, and that is:

$$ V_s(x) = \frac 1 2 k x^2 $$

The time coordinate can be define to that $t=0$ is when the mass hits the spring, or when it reaches equilibrium, or when the whole things starts. I like the penultimate, but will use the last.

So, at $t=0$:

$$ x(0) = 0.45\,{\rm m} \equiv x_0 $$

The total energy is (remember, the spring potential is "off"):

$$ E_0 = T(t=0) + V(t=0) = V_g(x_0) = mgx_0 $$

The time it takes to hit the spring, $t_1$, is a basic formula from:

$$ x_0 = \frac 1 2 g t_1^2 $$

so:

$$ t_1 = \sqrt{\frac{2x_0}{g}}$$

with (energy conservation):

$$ E_1 = T_1 + V_1 = E_0 $$

$$ \frac 1 2 mv^2_1 + 0 = mgx_0 $$

with

$$ v_1 = gt_1 = \sqrt{2gx_0} $$

Next, is calculate the equilibrium position, $x'$, which is the solution to:

$$ \frac{dU(x)}{dx} = 0 $$

$$ mg + kx' = 0 $$

$$ x' = -\frac{mg} k $$

Note that the energy of the compressed spring at equilibrium is:

$$ V_k(x') = \frac 1 2 k \frac{(mg)^2}{k^2} = \frac 1 2 \frac{(mg)^2}{k}$$

While the gravitational potential difference from $x=0$ is

$$ V_g(x') = mgx' = \frac{(mg)^2}{k} = 2V_k(x') $$

That $V_g/V_k = 2$ in equilibrium comes up a lot in physics problem (remember it). If you consider the harmonic oscillator as your bound system and "gravity" to be an external perturbing field, you'll see this in all kinds of systems, such as a polarizable atom in an external uniform electric field.

From here: write the equations of motion:

$$ ma - F = 0$$ $$ m\ddot x - kx - mg = 0 $$

and transform to $y = x-x' = x +\frac{mg} k $:

$$ m\ddot y - k(y-\frac{mg} k) + - mg = 0 $$ $$ m\ddot y - ky+\frac{kmg} k + - mg = 0 $$ $$ m\ddot y - ky = 0 $$

Notice how gravity has completely dropped out of the problem. This is unique (afaik) to the SHO.

So now solve that equation with the correct initial conditions: You know $y(t_1) = +|x'|$ and $\dot y(t_1) = v_1$.

So to reiterate: do not use real numbers until the end. Use physical quantities as variables, with clear and obvious names. Mark significant events with subscripts. At each step, think about "what conserved quantity do I know at that this step?", and "how does it get me to the next step?".

Finally: think about energy and global concerns, over force and local concerns. For instance, I didn't balance the spring/gravity forces to find equilibrium; rather, I only considered the total potential energy.