We need to know the speed of light, which is:

\(\displaystyle c=3.0*10^8\frac{m}{s}\)

We can use this to calculate the wavelength of any electromagnetic wave if we know the frequency, using the equation:

\(\displaystyle \lambda = \frac{v}{f}\)

If you're unsure of how to come about this equation, you should be able to derive it by looking at your units. Use our given values to solve for the wavelength:

\(\displaystyle \lambda = \frac{3.0*10^8\frac{m}{s}}{92.9*10^6 s^{-1}} = 3.3 m\)

Since the wave has a frequency of 30Hz, we know it completes 30 cycles every second. As its velocity is \(\displaystyle 10\frac{m}{s}\), we also know that in that second, the wave has moved 10m. Dividing the total distance, 10m, by the number of cycles, 30, we get the number of meters travelled in each cycle which is the wavelength:

\(\displaystyle \frac{10m}{30}=\frac{1}{3}m\)

Speed is frequency times wavelength. So the frequency must be speed divided by wavelength. In this case:

\(\displaystyle \lambda=\frac{15}{0.2}=75Hz\) 

This makes sense as in one second, the wave will have travelled 15m. Each meter is 5 cycles, so over the distance the wave has travelled in one second, it has completed \(\displaystyle 5*15=75\) cycles.

A trigonometric equation given by the following formula

\(\displaystyle y(t)=Asin(B(t-C))+D\)

Here, \(\displaystyle A\) is given in meters, \(\displaystyle B\) is given in Hertz, \(\displaystyle t\) is given in seconds, and \(\displaystyle D\) is given in meters. The period is given by:

\(\displaystyle period=\frac{2 \pi}{B}\)

In our case:

\(\displaystyle period= \frac{2 \pi}{2 \pi s^{-1}}=1s\)

Radio waves travel at the speed of light \(\displaystyle c\). The relationship between wavelength, frequency, and wave speed is

\(\displaystyle c=\lambda f\), where \(\displaystyle \lambda\) is the wavelength and \(\displaystyle f\) is the frequency. 

In our case, 

\(\displaystyle f=103.5\times10^6\:s^{-1}\). 

\(\displaystyle c=3.0\times10^8\:\frac{m}{s}\)

Therefore,

\(\displaystyle \frac{c}{f}=\lambda=\frac{3\times10^8}{1.035\times10^8}=2.90\:m\)

Amplitude refers to the strength of a wave and has no relation to its frequency. Frequency would only affect the period and phase, but not the amplitude. 

The relationship between wavelength and velocity is given by the equation;

\(\displaystyle v=f\lambda\)

The question gives us the velocity of the wave and its frequency. Using these values, we can solve for the wavelength.

\(\displaystyle 50\frac{m}{s}=(20Hz)\lambda\)

\(\displaystyle \lambda=\frac{50\frac{m}{s}}{20Hz}=2.5m\)

For a pendulum on a string, the period at which it oscillates is:

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

Period is the reciprocal of frequency.

\(\displaystyle T = \frac{1}{f}\)

Calculate the frequency of oscillation.

\(\displaystyle f = \frac{1}{2\pi}\sqrt{\frac{g}{l}}\)

The only parameter that the frequency depends on is the length \(\displaystyle l\). Decreasing length increases frequency. 

In this question, we're given the wavelength of light in a vaccuum. Since we know that the speed of light in a vaccuum is equal to \(\displaystyle 3.0\cdot10^{8}\: \frac{m}{s}\), we can first calculate the frequency of the light, and then use this value to calculate its period.

To begin with, we'll need to make use of the following equation to solve for frequency.

\(\displaystyle c=f\lambda\)

\(\displaystyle f=\frac{c}{\lambda }=\frac{3.0\cdot10^{8}\: \frac{m}{s}}{450\cdot 10^{-9}\: m}=6.67\cdot10^{14}\:s^{-1}\)

Then, using this value, we can calculate the period by noting that the period is the inverse of frequency.

\(\displaystyle T=\frac{1}{f}=\frac{1}{6.67\cdot10^{14}\:s^{-1}}=1.5\cdot10^{-15}\:s\)

And for completion's sake, it is worth noting that the period refers to the amount of time it takes for one wavelength to pass, which in this case is a very, very small fraction of a second!

In order to calculate the period \(\displaystyle T\), we need to find the frequency, \(\displaystyle f\), at which the waves hit the pier. We are told that \(\displaystyle 25\) waves hit within a \(\displaystyle 60s\) window. Therefore,

\(\displaystyle f=\frac{25}{60s}=0.417\frac{1}{s}\). We can then calculate the period \(\displaystyle T\) from this information.

\(\displaystyle T=\frac{1}{f}=\frac{1}{0.417\frac{1}{s}}=2.4s\)