The period of a pendulum is given by the formula:

 is the length of the pendulum arm or string, and  is the acceleration due to gravity. Gravity is constant, suggesting that the only way to manipulate the period of the pendulum is by adjusting the length of the string. Doubling the length of the pendulum arm would therefore increase the period. Since mass does not appear in the equation, it has no effect on the period of a pendulum.

Mechanical energy is the sum of potential energy and kinetic energy. Mechanical energy is conserved, unless there is intervention by a non-conservative force (such as friction).

The total energy in the system is determined by how far the pendulum is raised before release, based on the initial potential energy.

The potential energy is all converted to kinetic energy as the pendulum swings through the bottom, or mid-point of its excursion.

Once the pendulum is set in motion, the mechanical energy is constant in a frictionless system.

Hooke’s law, which is applicable to simple harmonic motion, states the relationship between force (F) and displacement (d).

k is equal to the spring constant. The ratio of F (force) to x (displacement) will be equal to the magnitude of k. In our set-up, the force is equal to the weight of the pendulum, so the ratio of weight to displacement is equal to the spring constant. This is true of all pendulums, and is given by the equation .

This problem covers conservation of energy in the form of a pendulum:

If the initial state is when the pendulum is at its highest point, and the final state is when the pendulum is at its lowest state, we can rewrite:

Substituting in our expressions:

Rearranging for final velocity:

We can calculate the height using the maximum angle the pendulum makes to the vertical. At this point, the pendulum covers a vertical distance of:

Therefore, the height above the lowest point is:

Now, we just need to find out what the net downward acceleration is. Normally gravity is 10, but the elevator is accelerating upward at a rate of . Therefore the percieved gravitational acceleration is .

Plugging in all of our values:

This problem focuses on understanding simple harmonic motion. The formula to identify the frequency, or cycles per second, that the mass moves is given below, where  is the spring constant of the spring and  is the mass of the object attached.

Notice that the frequency does not depend on how far the mass is displaced originally. If the mass is displaced more, it will simply be moving faster when it goes through the equilibrium position. It will not, however, move through the equilibrium position any more frequently than it would if it had been displaced less. 

Increasing the mass of the object attached would increase the denominator in the formula, and would therefore decrease the frequency of the harmonic motion.

Increasing the spring constant, , would increase the numerator of the formula, which would result in an increase in the frequency on the motion. The spring constant is directly proportional to the stiffness of the spring; thus, increasing stiffness will increase frequency.

The maximum force that the spring will exert on the mass is at maximum displacement from the equilibrium position. This can be seen with the formula for spring force.

 is the displacement and  is the spring constant. The reason that the right side of the formula is negative is because the force due to the spring is in the opposite direction of the displacement. The spring constant remains the same throughout the motion of the mass, therefore the maximum force will be generated when  (displacement) is greatest.

The equation best suited to solve this question is , where is the frequency of the motion, is the spring constant, and is the mass.

Two approaches can be used to solve for the period.

Since  , we can first solve for the frequency, and then solve for .  The second method is to rearrange the equation to directly solve for : . Either method can be used; it is simply a matter of preference.

Using the first method, we can plug in the given values from the question.

Now, we need to take the reciprocal of the frequency in order to solve for the period.

Since the period is the inverse of the frequency, the answer is .

Looking at the equation, we can see that the distance the spring is stretched or compressed has no effect on the period or frequency.

Hooke's law is written as , where k is the spring constant, and  is the change in position of the spring.

Since this is the upward force that will cancel out the downward force of the plant on the spring, we can set the forces equal to each other.

Hooke's law states the formula for the force of a spring:

The distorting force on a spring, whether lengthening or shortening, is opposed by the spring, hence the negative sign. In our question, the distorting force is caused by gravity acting on the mass, allowing us to set the force of the spring equal and opposite to the force fo gravity on the mass.

Use our given values for the spring constant, mass, and acceleration of gravity to find the resultant displacement.

Hanging the mass from the spring will cause it to stretch approximately .

There are a few steps to this problem. First we need to know the tension on the spring. From there we can calculate the spring constant, knowing the potential energy it is storing.

Let's draw in all of the forces to help visualize the system:

If the system is at equilibrum, then all of the forces must cancel out. Separating the forces into their components we get:

x-components:

y-components:

We can solve these equations simultaneously for T1 and T2. Since we only need to know T2, let's rearrange the first equation for T1:

Substituting this into the second equation, we get:

Rearranging for T2:

Using our given values, we can solve for T2:

This is the tension of the spring. There are two equatons that we can write out for the spring:

The only variables we don't know are  and , so we can solve these two equations simultaneously. We want , so let's rearrange the first equation for :

Substituting this into the second equation, we get:

Rearranging for :

We know the force and potential energy, allowing us to solve: