This question deals with conservation of energy in the form of a pendulum. The equation for conservation of energy is:

\(\displaystyle E = U_i + K_i = U_f + K_f\)

According to the problem statement, there is no initial kinetic energy and no final potential energy. The equation becomes:

\(\displaystyle U_i = K_f\)

Subsituting in the expressions for potential and kinetic energy, we get:

\(\displaystyle mgh = \frac{1}{2}mv_f^2\)

We can eliminate mass to get:

\(\displaystyle gh = \frac{1}{2}v_f^2\)

Rearranging for final velocity, we get:

\(\displaystyle v_f = \sqrt{2gh}\)

In order to solve for the velocity, we need to find the initial height of the ball.

The following diagram will help visualize the system:

From this, we can write:

\(\displaystyle 4 = d + h\)

\(\displaystyle h = 4 - d\)

Using the length of string and the angle it's held at, we can solve for \(\displaystyle d\):

\(\displaystyle d = 4sin(30^{\circ}) = 2\)

\(\displaystyle h=4-d=4-2=2\)

Now that we have all of our information, we can solve for the final velocity:

\(\displaystyle v_f = \sqrt{2\cdot 10\cdot 2} = \sqrt{40} = 6.3 \frac{m}{s}\)

We need to know how to calculate the period of a pendulum to solve this problem. The formula for period is:

\(\displaystyle P = 2\pi \sqrt {\frac{L}{g}}\)

In the problem, we are only changing the length of the string. Therefore, we can rewrite the equation for each scenario:

\(\displaystyle P_1 = \frac{2\pi}{\sqrt{g}} \sqrt {{L_1}}\)

\(\displaystyle P_2 = \frac{2\pi}{\sqrt{g}} \sqrt {{L_2}}\)

Dividng one expression by the other, we get a ratio:

\(\displaystyle \frac{P_1}{P_2} = \sqrt{\frac{L_1}{L_2}}\)

We know that \(\displaystyle L_2=4L_1\), so we can rewrite the expression as:

\(\displaystyle \frac{P_1}{P_2} = \sqrt{\frac{L_1}{4L_1}} = \sqrt{\frac{1}{4}}=\frac{1}{2}\)

Rearranging for P2, we get:

\(\displaystyle P_2 = 2P_1 = 2(5s) = 10s\)

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

\(\displaystyle T=2\pi \sqrt{\frac{L}{g}}\)

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:

\(\displaystyle T_1 = 2\pi\sqrt{\frac{L}{g_1}}\)

\(\displaystyle T_2 = 2\pi\sqrt{\frac{L}{g_2}}\)

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:

\(\displaystyle \frac{T_2}{T_1} = \frac{2\pi\sqrt{\frac{L}{g_2}}}{2\pi\sqrt\frac{L}{g_1}}\)

Canceling out the constants and rearranging, we get:

\(\displaystyle \frac{T_2}{T_1} = \sqrt\frac{g_1}{g_2}\)

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:

\(\displaystyle v_f^2 = v_i^2 + 2a\Delta x\)

Removing initial velocity and rearranging for acceleration, we get:

\(\displaystyle a = \frac{v_f^2}{2\Delta x}\)

Plugging in our values:

\(\displaystyle a = \frac{144\frac{m^2}{s^2}}{2\cdot20m} = 3.6 \frac{m}{s^2}\)

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

\(\displaystyle T_2 = T_1 \sqrt{\frac{g_1}{g_2}}= (3s)\sqrt{\frac{10\frac{m}{s^2}}{3.6\frac{m}{s^2}}}\)

\(\displaystyle T_2 = 5 s\)

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

\(\displaystyle T = 2\pi \sqrt{{\frac{L}{g}}}\), which does not include mass.

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

 \(\displaystyle T= \frac{2\pi }{\sqrt{\frac{g}{L}}}\)

Substituting our values, we obtain:

\(\displaystyle T= \frac{2\pi }{\sqrt{\frac{10}{10}}} = \frac{2\pi}{1}\approx6.3\)

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, \(\displaystyle g\).  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 \(\displaystyle \sqrt{2}=1.41\). Neither of the other parameters (amplitude, phase) change.

The frequency of a pendulum is given by:

\(\displaystyle \omega=\sqrt{\frac{g}{L}}\)

Where \(\displaystyle L\) is the length of the pendulum and \(\displaystyle g\) is the gravity constant. Notice how the frequency is independent of mass.

Plugging in values:

\(\displaystyle \omega=\sqrt{\frac{9.8}{2.5}}\)

\(\displaystyle \omega=\frac{1.98}{s}\)

The frequency of a pendulum is given by:

\(\displaystyle \omega=\sqrt{\frac{g}{L}}\)

Where \(\displaystyle L\) is the length of the pendulum and \(\displaystyle g\) 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: 

\(\displaystyle T=2\pi \sqrt{\frac{L}{g}}\)

Where \(\displaystyle T\) is the period of the pendulum, \(\displaystyle L\) is the length of the pendulum, and \(\displaystyle g\) is the gravitational constant of the planet we are on. Thus on Earth, the period \(\displaystyle T_e\) is given by:\(\displaystyle T_e=2\pi\sqrt{\frac{L}{g_e}}\)

With \(\displaystyle g_e\) being Earth's gravitational constant. The period on the Moon is given by: 

\(\displaystyle T_m=2\pi\sqrt{\frac{L}{g_m}}\)

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

\(\displaystyle g_m=\frac{g_e}{6}\) 

Plug this value into the Moon pendulum equation: 

\(\displaystyle T_m=2\pi\sqrt{\frac{L}{\frac{g_e}{6}}}=2\pi\sqrt{\frac{6L}{g}}=2\pi\sqrt{\frac{L}{g_e}}\sqrt{6}\)

Since \(\displaystyle T_e=2\pi\sqrt{\frac{L}{g_e}}\),

Substituting this into the above expression gives us \(\displaystyle T_m=T_e\sqrt{6}\)