We need to know the formula for the period of a pendulum to solve this problem:

We aren't given the length of the pendulum, but that's ok. We could solve for it, but it would be an unnecessary step since the length remains constant.

We can write this formula for the pendulum when it is on the student's table and when it is falling:

1 denotes on the table and 2 denotes falling. The only thing that is different between the two states is the period and the gravity (technically the acceleration of the whole system, but this is the form in which you are most likely to see the formula). We can divide the two expressions to get a ratio:

Canceling out the constants and rearranging, we get:

We know g1; it's simply 10. However, we need to calculate g2, which is the rate at which the pendulum and balloon are accelerating toward the ground. We are given enough information to use the following formula to determine this:

Removing initial velocity and rearranging for acceleration, we get:

Plugging in our values:

This is our g2. We now have all of the values to solve for T2:

The mass of a pendulum has no effect on its period. The equation for the period of a pendulum is

, which does not include mass.

The period of a pendulum is given by the following formula:

Substituting our values, we obtain:

Roughly 6.3 seconds is the time it takes for the pendulum to complete one period.

The period and frequency of a pendulum depend only on its length and the gravity force constant, .  Changing the mass of the pendulum does not affect the frequency, and since the student released the new pendulum from the same displacement as the old, the amplitude and phase remain the same, and the equation of motion is the same for both pendula.

Doubling the length of a pendulum increases the period, so it decreases the frequency of the pendulum. The frequency depends upon the square root of the length, so the frequency decreases by a factor of . Neither of the other parameters (amplitude, phase) change.

The frequency of a pendulum is given by:

Where is the length of the pendulum and is the gravity constant. Notice how the frequency is independent of mass.

Plugging in values:

The frequency of a pendulum is given by:

Where is the length of the pendulum and is the gravity constant. The frequency is independent of mass. Thus, adding mass will have no effect.

The period of a simple pendulum is given by: 

Where  is the period of the pendulum,  is the length of the pendulum, and  is the gravitational constant of the planet we are on. Thus on Earth, the period  is given by:

With  being Earth's gravitational constant. The period on the Moon is given by: 

With  being the Moon's gravitational constant. Since the Moon's gravity is 6 times weaker than that of Earth's, we have: 

Plug this value into the Moon pendulum equation: 

Since ,

Substituting this into the above expression gives us 

For this question, we're presented with a scenario in which a pendulum is swinging back and forth. Thus, we know that this pendulum is an example of simple harmonic motion. As the pendulum swings back and forth, a number of its variables change in a cyclical fashion.

First, let's take a look at potential energy. When the pendulum is at the bottom of its trajectory, its potential energy will also be at its minimum. This is because the height of the mass attached to the pendulum is lowest at this point. We can show this with the following expression, where the  term is at a minimum.

Next, let's consider tangential acceleration. When the pendulum reaches its highest point, it will briefly be at rest for a very short instant. At this highest point, the pendulum also has its greatest amount of potential energy. When the pendulum begins to fall, the force due to gravity causes the pendulum to fall. However, it's important to realize that the force of gravity acts on the pendulum's mass in two ways. One way is tangentially, in which the force acts along the direction of the mass's motion. The other way is radially to the mass's direction of motion; in other words, along the pendulum's string. This can be shown with a diagram as follows.

As can be seen in the above diagram, the tangential acceleration is represented by the following expression.

Thus, as the pendulum swings to its lowest point, the value of  approaches zero. As it does this, the tangential acceleration also approaches zero.

Both the frequency and the period of the pendulum's harmonic motion is in no way related to the lowest point of the pendulum's path.

Finally, let's consider kinetic energy. We've already noted how the pendulum's potential energy is at a maximum at its highest point. As the pendulum falls to its lowest point, its potential energy is converted into kinetic energy. This is because as the pendulum falls to its lowest point, it speeds up more and more. Thus, at its lowest point, the pendulum has its kinetic energy at a maximum.

The expression for the period of a pendulum is:

Therefore, the period of a pendulum is proportional to the square root of the length of the pendulum (assuming they are both on earth, or the same planetary body). Thus, the pendulum with the longer length will have the longer period.