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,

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

The expression for the frequency of a spring:

$\displaystyle f = \frac{w}{2\pi}$

Therefore, we can say:

$\displaystyle \frac{f_1}{f_2}=\frac{\left ( \frac{w_1}{2\pi}\right )}{\left ( \frac{w_2}{2\pi}\right )}$

$\displaystyle \frac{f_1}{f_2}=\frac{w_1}{w_2}$

Where:

$\displaystyle w = \sqrt{\frac{k}{m}}$

Substituting this into our expression, we get:

$\displaystyle \frac{f_1}{f_2}=\frac{\sqrt{\frac{k}{m_1}}}{\sqrt{\frac{k}{m_2}}}$

$\displaystyle \frac{f_1}{f_2}= \sqrt{\frac{m_2}{m_1}}$

Taking the inverse of both sides:

$\displaystyle \frac{f_2}{f_1}= \sqrt{\frac{m_1}{m_2}}$

Rearranging for final frequency:

$\displaystyle f_2= f_1\sqrt{\frac{m_1}{m_2}}$

From the problem statement, we know that:

$\displaystyle m_2 = 2m_1$

substituting this into the expression, we get:

$\displaystyle f_2 = f_1\sqrt{\frac{m_1}{2m_2}} = f_1\sqrt{\frac{1}{2}}$

Therefore, the frequency was changed by a factor of $\displaystyle \sqrt{\frac{1}{2}}$

We only need one expression to solve this problem:

$\displaystyle f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$

Now we just need to rearrange for the length of the pendulum:

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

$\displaystyle \frac{g}{L} = (2\pi f)^2$

$\displaystyle L = \frac{g}{(2\pi f)^2}$

We have values for each of these variables, so time to plug and chug:

$\displaystyle L = \frac{\left ( 10\frac{m}{s^2}\right )}{\left ( 2\pi 0.4\frac{1}{s}\right )^2}$

$\displaystyle L = 1.6m$

Since we are told that this is a simple pendulum, we can use the follow expression for period:

$\displaystyle T_1=2\pi \sqrt{\frac{L_1}{g_1}}$

$\displaystyle T_2=2\pi \sqrt{\frac{L_2}{g_2}}$

Now let's divide scenario 2 by scenario 1:

$\displaystyle \frac{T_2}{T_1}= \sqrt{\frac{L_2g_1}{L_1g_2}}$

From the problem statement, we know that:

$\displaystyle L_2 = 2L_1$

So let's plug that in:

$\displaystyle \frac{T_2}{T_1}= \sqrt{\frac{2L_1g_1}{L_1g_2}} = \sqrt{\frac{2g_1}{g_2}}$

Then plugging in our values for g:

$\displaystyle \frac{T_2}{T_1} = \sqrt{\frac{2\left ( 10\frac{m}{s^2}\right )}{\left ( 1.6\frac{m}{s^2}\right )}}$

$\displaystyle \frac{T_2}{T_1} = 3.5$

An object has the maximum potential energy the furthest from its equilibrium point (at the turnaround point). So it at the equilibrium position it would have the minimum potential energy. If it is undergoing simple harmonic motion, it would have the maximum kinetic energy as it passes through the equilibrium position because it is returning from the stretched position (spring example) where it gathered energy. The same is true for other objects undergoing this motion.

It's impossible because the period of a simple pendulum doesn't depend on the mass of the bob. Because of this, we have no way to determine the mass from the period.

In these types of problems, it is always advantageous to recognize the format of the equation. In trigonometric functions, the period is always given by $\displaystyle \frac{2\pi}{B}$, when the function is written as $\displaystyle Asin(Bt+C)+D$. Since frequency is the reciprocal of the period, we will need to flip the fraction.

$\displaystyle B=4\pi$

$\displaystyle P=\frac{2\pi}{B}=\frac{2\pi}{4\pi}=\frac{1}{2}$

$\displaystyle Frequency=\frac{1}{P}=\frac{1}{\frac{1}{2}}=2Hz$

In trigonometric functions, the period is always given by $\displaystyle \frac{2\pi}{B}$, when the function is written as $\displaystyle Asin(Bt+C)+D$. Once, we determine our $\displaystyle B$ value, we are halfway to the solution!

$\displaystyle B=4\pi$

$\displaystyle P=\frac{2\pi}{B}=\frac{2\pi}{4\pi}=\frac{1}{2}s$

Using conservation of energy:

$\displaystyle KE_1+EPE_1=KE_2+EPE_2$

$\displaystyle .5mv_1^2+.5kx_1^2=.5mv_2^2+.5kx^2$

Plugging in values:

$\displaystyle .5*.050*2^2+.5k(0)^2=.5*.050(0)^2+.5k*.05^2$

Solving for $\displaystyle k$

$\displaystyle k=\frac{80N}{m}$

$\displaystyle \omega=\sqrt{\frac{k}{m}}$

Plugging in values:

$\displaystyle \omega=\frac{40 rad}{s}$