The base equation for position when undergoing simple harmonic motion is:

First, solve for the phase constant.

Plug all the variables into the equation and solve.

Recall Hooke's Law, which states:

Here,  is the force exerted by the spring,  is the spring constant, and  is the displacement from the spring's rest position. This equation tells us that the force exerted is directly proportional to the displacement. We don't need to solve for  to determine the magnitude of the force on the spring stretched 10m. We can instead come up with a proportionality such that:

Here,  and  are forces applied on the string and  and  are the displacements of the spring from its respect position respectively. We assume that a stretched spring will have a positive displacement, whereas a shrunken spring will have a negative displacement. However, since we're looking for the magnitude of the force, regardless of direction, the direction of the displacement doesn't matter. Therefore, we can write the proportion as:

In our case:

Hooke's law states that the spring force is equal to the product of the spring constant and the displacement of the spring:

The force is negative because it acts in the direction opposite of the displacement from the equilibrium position (i.e. when we stretch we do so in the positive direction). We are given the force and the displacement, so we just solve for k:

This question is giving us the amount of force needed to displace an object attached to a spring, and is asking us to calculate the spring constant. Thus, we will need to make use of Hooke's law.

The above equation, Hooke's law, tells us that the restoring force of the spring is related to the displacement of the attached object. It's important to note that in the question stem, we are told that an external force of  is needed to displace this object. Thus, the restoring force of the spring, due to Newton's third law, has the same magniture of the applied force but in the opposite direction. Thus, the restoring force of the spring is equal to . Plugging this value into Hooke's law, as well as the displacement of , yields:

Where is the spring constant

is the compression of the spring

is the mass of the object.

is the gravity constant, which will be treated as a negative number.

Solve for :

Plug in values:

Where and are the respective spring constants, is the stretch length, and is the gravity constant, which is a negative as the vector is pointing down.

Since

Substitute and plug in values:

Solving for :

Determine the net forces on the submarine:

Plug in values:

Determine what forces are acting on the submarine:

Plug in values

*Note: Acceleration due to gravity is going down so it is a negative

Solve for :

Each spring will be subject to the same force, and since they have the same spring constant, stretch the same amount. Thus:

Total stretch:

Stretch of one spring:

Use Hooke's law:

The force will be equal to the force of gravity on the mass:

Solve for :

The force of the spring in relationship to strain is independent of direction. Thus, the same force pulling on the spring would result in an equal amount of length change, albeit by stretching instead of compressing.

In this problem the "spring within a spring" can be treated as a single spring.

Use Hooke's law:

Plug in values:

Solve for .

Again use Hooke's Law:

Plug in values:

Solve for