To solve this problem, we must use our knowledge of the relationship between angular frequency, the spring constant, and the mass.

We can then use this, and the derived maximum speed of the object in terms of the amplitude and angular frequency, to find the answer.

Doubling the initial distance the weight is displaced from equilibrium doubles the amplitude, but does not change the period or the phase of the oscillator. So in the general equation of an oscillator:

, only the  term changes, and it doubles.

The amplitude does not change since the student pulls the weight the same displacement. Changing the spring constant changes the frequency of the oscillator. Since the period decreases, the frequency increases, but it does not double. The frequency and the period depend on the square root of the spring constant, so the frequency increases by . In the equation of motion for an oscillator,

, only the frequency , changes.

Since there are two forces acting on the object, and it isn't moving, there is an equality of the two forces. This means that:

Equivalently:

Where  is the spring constant,  is the displacement of the spring,  is the equilibrium position,  is mass of the object and  is the gravitational constant given as 

Since we know that the mass on the spring spring is displaced , the mass is , and we know the gravitational constant we plug in and solve for the spring constant.

Use the equation:

for a mass on a horizontal spring (so no gravity in the direction of motion)

The stretch length will have no effect.

Plug in values:

We know that when the block is at its maximum displacement, it is not moving and therefore has no kinetic energy. Also, we know that the block has its maximum kinetic energy (thus velocity) when it is passing through the springs equilibrium and therefore has no potential energy. Therefore, we can say:

Where the potential energy of a spring is:

And the expression for kinetic energy is:

Substituting in these expressions, we get:

Canceling out the fractions:

No rearranging for mass, we get:

We have all of these values, so we can solve the problem:

The expression for the period of a spring in simple harmonic motion is:

We have the spring constant, so we just need to calculate the mass of the block. For a spring in simple harmonic motion, we know that the maxmium potential energy is equal to the maximum kinetic energy. Therefore we can say:

Rearranging for mass, we get:

Plugging in our values, we get:

Now we can use the expression for period to solve the equation:

Plugging in our values:

We can use the expression for conservation of energy to solve this problem:

We can eliminate initial kinetic energy (block initially at rest) and final potential energy (block at equilibrium of spring) to get:

Substituting in expressions for each of these, we get:

Where initial height is simply the displacement of the spring:

Multiplying both sides of the expression by 2 and rearranging for final velocity, we get:

Plugging in values for each of these variables, we get:

The spring constant is equal to .

Using conservation of energy:

While at the maximum value of stretch (the amplitude), the velocity will be zero. While at the maximum velocity, the stretch will be zero.

Plugging in values:

Solving for 

Plugging in values: