The mass has no bearing on the relationship between frequency and period. This relationship is given by the equation:

$\displaystyle f = \frac{1}{T}$

Given the period, the frequency will be equal to its reciprocal.

$\displaystyle f = \frac{1}{T}$

$\displaystyle f = \frac{1}{3s}$

$\displaystyle f=0.33Hz$

To begin we want to determine the original length of the pendulum.

The equation to determine the period of pendulum is

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

We can rearrange this equation to solve for length ($\displaystyle L$).

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

$\displaystyle \frac{T^2}{4\pi ^2} = \frac{L}{g}$

$\displaystyle \frac{T^2g}{4\pi ^2} = L$

We can now solve for the length knowing that the period of the pendulum is 2 seconds.

$\displaystyle \frac{2^2(9.8)}{4\pi ^2} = L$

$\displaystyle 0.99m = L$

Now we can use the same equation and instead substitute in the value for the acceleration due to gravity on Mars.

$\displaystyle T = 2\pi \frac{L}{g}$

$\displaystyle T = 2\pi \sqrt{\frac{0.99}{(0.37)(9.8)}}$

$\displaystyle T = 3.3s$

The equation to determine the period of pendulum is

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

We can rearrange this equation to solve for length ($\displaystyle L$).

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

$\displaystyle \frac{T^{2}}{4\pi ^{2}}=\frac{L}{g}$

$\displaystyle \frac{T^2g}{4\pi ^2} = L$

We can now solve for the length knowing that the period of the pendulum is 2 seconds.

$\displaystyle \frac{2^2(9.8)}{4\pi ^2} = L$

$\displaystyle 0.99m = L$

The period is how long it takes to complete one oscillation. In this case, a complete oscillation would take the object $\displaystyle 0.21m$ away from the center (the amplitude), travel back to the center, travel below the equilibrium point $\displaystyle 0.21m$ away, and then travel back to the center. Therefore the total distance traveled is $\displaystyle 4$ times the amplitude as it must travel away, back, below, and back again. Therefore the total distance traveled is $\displaystyle 0.84m$.

The period of a simple pendulum on the free-falling is infinite because at a time of free-falling elevator the gravitational constant will equal zero which leads to an increase in the value in the numerator to infinite. Thus, the period of time in a free-falling elevator is infinite.

The best way to solve this problem is by using energy. Notice that the spring on its own is stationary. That means its initial total energy at that moment is zero. When the mass is attached, the spring stretches out, giving it spring potential energy (PEspring).

Where does that energy come from? The only place it can come from is the addition of the mass. Since the system is vertical, this mass will have gravitational potential energy. 

Use the law of conservation of energy to set these two energies equal to each other:

$\displaystyle PE_{mass}=PE_{spring}$

$\displaystyle mg\delta y=\frac{1}{2}k\delta y^2$

We are trying to solve for displacement, and now we have an equation in terms of our variable.

Start by diving both sides by Δy to get rid of the Δy2 on the right side of the equation.

$\displaystyle mg=\frac{1}{2}k\delta y$

We are given values for the spring constant, the mass, and gravity. Using these values will allow use to solve for the displacement.

$\displaystyle mg=\frac{1}{2}k\delta y$

$\displaystyle 8kg(-9.8m/s^2)=\frac{1}{2}(11.1N/m)\delta y$

$\displaystyle -78.4N =5.55N/m\delta y$

$\displaystyle \frac{-78.4N}{5.55Nm} =\delta y$

$\displaystyle -14.13m =\delta y$

Note that the displacement will be negative because the spring is stretched in the downward direction due to gravity.

If we're looking for the maximum velocity, that will happen when all the energy in the system is kinetic energy.

We can use the law of conservation of energy to see $\displaystyle PE_i=KE_f$. So, if we can find the initial potential energy, we can find the final kinetic energy, and use that to find the mass's final velocity.

The formula for spring potential energy is:

$\displaystyle PE = \frac{1}{2}kx^2$

Plug in our given values and solve:

$\displaystyle PE = \frac{1}{2}(200N/m)(0.2m)^2$

$\displaystyle PE = 4J$

 The formula for kinetic energy is:

$\displaystyle KE=\frac{1}{2}mv^2$

Since $\displaystyle PE_i = KE_f$, that means that $\displaystyle KE = AJ$.

We can plug in that information to the formula for kinetic energy to solve for the maximum velocity:

$\displaystyle KE=\frac{1}{2}mv^2$

$\displaystyle 4J=\frac{1}{2}(0.56kg)v^2$

$\displaystyle 14.29m^2/s^2 = v^2$

$\displaystyle v=3.78m/s$

Conservation of energy dictates that the total mechanical energy will remain constant. Initially, the mass will not be moving and will be at its highest height. When released, it will begin to travel downward (lose potential energy) and gain velocity (gain kinetic energy). When the mass reaches the bottommost point in the swing, the potential energy will be at a minimum and the kinetic energy will be at a maximum. This point corresponds to the string being perpendicular to the ground.

We can start out by examining the energy in the pendulum.  At the top of the swing, there is gravitational potential energy.  At the bottom of the swing there is kinetic energy.  The law of conservation of energy states that these two values must be the same.

$\displaystyle GPE_{top} = KE_{bottom}$

We know the equations for gravitational potential energy and kinetic energy to be

 $\displaystyle GPE=mgh$

$\displaystyle KE=\frac{1}{2}mv^2$

We can set these equal to each other.

 $\displaystyle mgh = \frac{1}{2}mv^2$

 Since the mass is constant, it falls out of both sides of the equation.

 $\displaystyle gh = \frac{1}{2}v^2$

 Let’s rearrange and solve for v by itself.

 $\displaystyle 2gh = v^2$

 $\displaystyle v=\sqrt{2gh}$

 We don’t know the height of the pendulum at this point and need to get it in terms of the length of the pendulum.

If we knew the angle that the pendulum made with the vertical equilibrium point, we could determine how far off the ground the pendulum was.  We can create a triangle with the hypotenuse as the length of the string and the angle between the pendulum and the equilibrium point.

$\displaystyle Adjacent=Hypotenusecos(\theta )$

$\displaystyle Adjacent = Lcos(\theta )$

 This adjacent side can then be subtracted from the original length of the pendulum to determine the height off the ground.

$\displaystyle h=L-Lcos(\theta )$

 We can substitute this back into our equation

 $\displaystyle v=\sqrt{2g(L-Lcos(\theta ))}$

We will need to start at the end of the situation and work backwards in order to determine the velocity of the bullet.  At the very end, the pendulum with the bullet reaches its maximum height and therefore comes to a stop.  It has gravitational potential energy.  At the bottom of the pendulum right after the bullet collides with it, it has kinetic energy due to the velocity of the bullet.  With the law of conservation of energy we can set the kinetic energy of the pendulum right after the collision equal to the gravitational potential energy of the pendulum at the highest point.

To determine the height of the pendulum we will need to use trig and triangles to find the height.  We know that the pendulum makes a 30 degree angle with the equilibrium position at its maximum height.  The length of the pendulum is provided which is the hypotenuse of this triangle.  We need to find the adjacent side of this triangle.  We can use cosine to determine this.

$\displaystyle Adjacent = Hypotenuse cos(\theta)$

$\displaystyle Adjacent = 1m cos(30)$

$\displaystyle Adjacent = 0.866m$

We can now subtract this value from the length of the pendulum to determine how high off the ground the pendulum is at its highest point.

$\displaystyle 1m-0.866m = 0.134m$

We can now set the kinetic energy of the pendulum right after the collision equal to the gravitational potential energy of the pendulum at the highest point.

$\displaystyle KE = GPE$

$\displaystyle \frac{1}{2}mv^2 = mgh$

The mass is the same throughout so it falls out of the equation.

$\displaystyle \frac{1}{2}v^2 = gh$

$\displaystyle v^2 = 2gh$

$\displaystyle v = \sqrt{2gh}$

$\displaystyle v = \sqrt{2(9.8m/s^2)(0.134m)}$

$\displaystyle v = 1.62m/s$

The pendulum with the bullet was moving $\displaystyle 1.62m/s$ after the collision.  We can now use momentum to determine the speed of the bullet before the collision.  Conservation of momentum states that the momentum before the collision must equal the momentum after the collision.

$\displaystyle m_1 v_{i1} + m_2 v_{i2} = m_1 v_{f1} + m_2 v_{f2}$

$\displaystyle m_1 = 0.25g = 0.00025kg$

$\displaystyle m_2 = 100g = 0.1kg$

They both move together after the collision

$\displaystyle v_{f1} = v_{f2} = 1.62m/s$

Since the pendulum was not moving at the beginning

$\displaystyle v_{i2} = 0m/s$

We can now plug in these values and solve for the missing piece.

$\displaystyle 0.00025kg v_{i1} + 0 = 0.00025kg(1.62m/s) + 0.1kg(1.62m/s)$

$\displaystyle 0.00025kg v_{i1} = .162405kgm/s$

$\displaystyle v_{i1} = 649.62m/s = 650m/s$