We need to know three things to solve this problem.

1. Formula for frequency of a spring in simple harmonic motion
2. The spring constant
3. Mass of object attached to spring

We're given 2 and 3, so we just need to know 1.

The formula for frequency is:

$\displaystyle f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Plugging in our values, we get:

$\displaystyle f = \frac{1}{2\pi}\sqrt{\frac{10\frac{N}{m}}{2kg}} = \frac{1}{2\pi}\sqrt{5\frac{1}{s^2}}$

$\displaystyle f = 0.356 Hz$

Now we need to convert Hertz (cycles per second) to cycles per minute.

To get to these units:

$\displaystyle f = 0.356 \frac{1}{s}(\frac{60s}{min}) = 21.4\ min^{-1}$

We simply need the expression for the period of a pendulum to solve this problem:

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

Rearranging for length, we get:

$\displaystyle l = g\left ( \frac{T}{2\pi}\right )^2$

We have all of these values, allowing us to solve:
$\displaystyle l = (10\frac{m}{s^2})\left ( \frac{5s}{2\pi}\right )^2 = 6.3m$

First, we will be using the equation for the period of a pendulum:

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

The only value we don't have is length. However, we can develop an expression for length from the given information.

$\displaystyle h_{max} = l-lsin(A_{min})$

The second term describes how close the pendulum gets to the height of the top of the pendulum. Therefore, we subtract this value from the lowest point of the pendulum to get the height relative to its lowest point. We can rewrite this as:

$\displaystyle h_{max}= l(1-sin(A_{min}))$

$\displaystyle l = \frac{h_{max}}{1-sin(A_{min})}$

Substituting this into the original expression for the period, we get:

$\displaystyle T = 2\pi\sqrt{ \frac{h_{max}}{g(1-sin(A_{min}))}}$

We can use our given values to solve:

$\displaystyle T = 2\pi \sqrt{ \frac{0.75m}{(10\frac{m}{s^2})(1-sin(30^{\circ}))}}$

$\displaystyle T = 2.43s$

For a mass-spring system undergoing simple harmonic motion, the frequency of the oscillations can be found using the equation

$\displaystyle f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

We were given the force constant (or spring constant), $\displaystyle k$, to be $\displaystyle 250 \:\frac{N}{m}$. The oscillating mass was also given to be 1.3 kg. So, plug these in to the equation and solve for frequency, $\displaystyle f$. The unit for frequency is Hertz, Hz.

$\displaystyle f=\frac{1}{2\pi }\sqrt{\frac{250\:\frac{N}{m}}{1.3\:kg}}=2.2\:Hz$

We were given the mass of the system as 300 g. First, we should convert this to kilograms. Since $\displaystyle 1\:kg = 1000\:g$, we can convert by

$\displaystyle m=300\:g\cdot \frac{1\:kg}{1000\:g}=0.3\:kg$

They also told us the period of the oscillations, which is 0.25 s.  We can use the following equation to solve for the force constant:

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

Rearrange the equation to isolate the variable $\displaystyle k$, and substitute in the known values to solve for $\displaystyle k$:

$\displaystyle k=m\left ( \frac{2\pi }{T}\right )^{2}=0.3\:kg\cdot \left ( \frac{2\pi }{0.25\:s} \right )^{2}=189.5\:\frac{N}{m}$

Now that we have the force constant, $\displaystyle k$, we can use this to help us find the amplitude of the oscillations. We were also given the energy of the system to be 3.0 J. The equation for the energy of a mass-spring oscillating system is:

$\displaystyle PE_{s}=\frac{1}{2}kx^{2}$ 

$\displaystyle PE_s$ is the total potential energy of the system. Now, plug in values and solve for $\displaystyle x$, which is the amplitude of the oscillations.

$\displaystyle x=\sqrt{\frac{2\cdot PE_s}{k}}$

$\displaystyle x=\sqrt{\frac{2\cdot 3.0\:J}{189.5\:\frac{N}{m}}}=0.18\:m$

The equation that relates the period of a pendulum and its length is

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

We know the period of the pendulum on Earth, which was given, to be 1.0 s. We also know that $\displaystyle g$ on Earth is $\displaystyle 9.8 \:\frac{m}{s^2}$.  So, we can find the length of the pendulum:

$\displaystyle l=g_{earth}\left (\frac{T_{earth}}{2\pi } \right )^{2}=9.8\:\frac{m}{s^2}\left ( \frac{1.0\:s}{2\pi } \right )^{2}=0.248\:m$

Now that we have the length of the pendulum, we can use this to calculate what the period of oscillation will be if the pendulum was on the Moon where the "g" is not the same as on Earth, but rather is $\displaystyle 1.6 \:\frac{m}{s^2}$.

$\displaystyle T_{moon}=2\pi\sqrt{\frac{l}{g_{moon}}}=2\pi\sqrt{\frac{0.248\:m}{1.6\:\frac{m}{s^2}}}=2.5\:s$

Equation for period of a pendulum:

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

Solve.

$\displaystyle 7s = 2\pi \sqrt{\frac{L}{10\frac{m}{s^2}}}$

$\displaystyle L=12.4m$

The equation of motion for an oscillator is:

$\displaystyle x(t)=Acos(\omega t+\phi)$

In words, the position as a function of time equals the amplitude times the cosine of the sum of the frequency (in radians per second) times time plus phase. Changing the mass of the weight only changes the frequency, but it depends upon the square root of the mass. Since increasing the mass increases the period, it decreases the frequency by the factor of $\displaystyle \sqrt{2}=1.41$.

First, we want to ignore the summation sign and the $\displaystyle c_n$ since those terms do not affect the wavelength at $\displaystyle n=3$. All we need to look at is the sine term. 

For $\displaystyle n=3$, we get the sinusoid:

$\displaystyle sin(\frac{3 \pi x}{l})$ 

Remember that wavelength $\displaystyle \lambda$ is given by 

$\displaystyle \lambda = \frac{2 \pi}{\frac{3\pi}{l}}= \frac{2l}{3}$

Period is given by: $\displaystyle \frac{1}{f}$ where $\displaystyle f$ is frequency. Therefore,

$\displaystyle period=\frac{1}{2Hz}=\frac{1}{2}s$