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.

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

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

$\displaystyle f=0.33Hz$

The formula for the potential energy in a spring is:

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

Use the given spring constant and displacement to solve for the stored energy.

$\displaystyle PE=\frac{1}{2}(1500\frac{N}{m})(0.87m)^2$

$\displaystyle PE=(750\frac{N}{m})(0.7569m^2)$

$\displaystyle PE=567.68J$

When it comes to the period of a spring, the velocity of the object has no effect.

The equation is $\displaystyle T=2\pi\sqrt{\frac{m}{k}}$.

The equation for the period of a spring in simple harmonic motion is:

$\displaystyle T=2\pi\sqrt{\frac{m}{k}}$

In this formula, $\displaystyle m$ is the mass and $\displaystyle k$ is the spring constant. The only two things we can adjust that can change the period, then, are the mass and spring constant. The length of the spring and the acceleration due to gravity are irrelevant.

If we increase the mass we get a larger numerator, which in turns will give us a larger period. If we decrease the mass we get a smaller numerator, which would give us a smaller period. If we use a higher spring constant we get a larger denominator, which also gives us a smaller period.

The correct answer is that the force acts in the opposite direction of the displacement.

The equation given is Hooke's law, which is used to determine spring force based on the spring constant, $\displaystyle k$, and the displacement of the spring, $\displaystyle x$.Negative and downward are arbitrary, and spring constants are always positive. If a spring is stretched or compressed in a given direction, the force of the spring will always act opposite the direction of the displacement in order to return the spring to the resting equilibrium position.

If no outside forces act upon the pendulum, it will continue to oscillate back to the original height of $\displaystyle 3m$. 

The proof of this is in the law of conservation of energy. At the top, the pendulum has all potential energy, which is given by the formula $\displaystyle PE=mgh$. As it swings, the potential energy is converted to kinetic energy until, at the bottommost point, there is only kinetic energy. It then changes direction and begins to rise again. When it rises to the maximum height on the other side, all of its kinetic energy will turn back into potential energy.

Mathematically, the initial and final potential energies are equal.

$\displaystyle PE_1=PE_2$

$\displaystyle mgh_1=mgh_2$

Notice the masses and gravity can cancel out on both sides, as neither of these will change. This leaves us with only height.

$\displaystyle h_1=h_2$

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.

The equation for the period of a pendulum is:

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

Notice that the material of the pendulum and the mass at the end do not enter into the equation at all. The only things that are capable of affecting the period are are the length of the pendulum and the acceleration due to gravity.

While changing the altitude of the pendulum will change the period, it will do it only slightly unless you take it miles above the earth's crust. Furthermore, decreasing the altitude of the pendulum will increase the force due to gravity, which will resulut in a decreased period. The best answer is to increase the length of the rope. This will increase the period of the pendulum.

The equation for spring potential energy is $\displaystyle PE=\frac{1}{2}kx^2$.

Plug in the given values for the distance and spring constant to solve for the potential energy.

$\displaystyle PE=(\frac{1}{2}) (200\frac{N}{m}) (- 0.5m)^2$

Remember, since the spring was compressed, it has a negative displacement. The resultant potential energy will be positive as, when released, the displacement will be along the positive horizontal axis.

$\displaystyle PE=(\frac{1}{2})(200\frac{N}{m}) (0.25m^2)$

$\displaystyle PE=\frac{1}{2}(50\frac{N\cdot m^2}{m})$

$\displaystyle PE=\frac{1}{2} (50Nm)$

$\displaystyle PE=25J$

For this problem, use Hooke's law:

$\displaystyle F=k\Delta x$

In this formula, $\displaystyle k$ is the spring constant, $\displaystyle \Delta x$ is the compression of the spring, and $\displaystyle F$ is the necessary force. We are given the values for the spring constant and the distance of compression. Using these terms, we can sovle for the force of the spring.

Plug in our given values and solve.

$\displaystyle F=k\Delta x$

$\displaystyle F=200\frac{N}{m}(-0.1m)$

$\displaystyle F=-20N$

Note that the force is negative because it is compressing the spring, pushing against the coil. When the force is released, the equal and opposite force of the spring will cause it to extend in the positive direction.