Longitudinal waves are waves whose particles travel parallel to the direction that the wave itself is traveling. Sound waves are another example of longitudinal waves.

The speed of the wave along the Slinky depends on the mass of the Slinky itself and the tension caused by stretching it. Since both of these things have not changed, the wave speed remains constant.

The wave speed is equal to the wavelength multiplied by the frequency.

$\displaystyle v = \lambda f$

Since she is moving her hand faster, the frequency has increased. Since the velocity has not changed, an increase in the frequency would decrease the wavelength.

Sound is often described in terms of the vibration of the molecules of the medium in which it travels (in other words, the displacement of the molecules). Sound can also be viewed from a pressure point of view because this variation in pressure is easier to measure. In compression, the pressure is higher because the molecules are closer together. In rarefaction, there is an expansion of molecules and, therefore, a lower pressure.

When you see a relationship like "times every second" or "once per hour," these are hints you are looking at a frequency. Frequency is, effectively, how often something happens. If it happens four times per second, then we know how often it happens. The units "per second" are equivalent to Hertz.

The relationship between frequency and period is $\displaystyle T=\frac{1}{f}$.

Since our given frequency was four beats per second, or $\displaystyle 4Hz$, we can solve for the period.

$\displaystyle T=\frac{1}{4Hz}=\frac{1}{4}s=0.25s$

This means that her heart beats once every 0.25 seconds.

The relationship between wavelength and frequency is given by the equation $\displaystyle \lambda =\frac{c}{f}$, where $\displaystyle \lambda$ is the wavelength, $\displaystyle c$ is the speed of light, and $\displaystyle f$ is frequency.

We are given the values for frequency and the speed of light, allowing us to solve for the wavelength.

$\displaystyle \lambda =\frac{c}{f}$

$\displaystyle \lambda=\frac{3*10^8\frac{m}{s}}{99,500,000Hz}$

$\displaystyle \lambda=3.02m$

The equation for velocity in terms of wavelength and frequency is $\displaystyle v=\lambda f$.

We are given the velocity and the wavelength. Using these values, we can solve for the frequency.

$\displaystyle v=\lambda f$

$\displaystyle 8.9\frac{m}{s}=6.7m* f$

$\displaystyle \frac{8.9\frac{m}{s}}{6.7m}=f$

$\displaystyle 1.33Hz=f$

The relationship between velocity, frequency, and wavelength is:

$\displaystyle v=f\lambda$

Plug in the given information to solve:

$\displaystyle v=f\lambda$

$\displaystyle 340.29\frac{m}{s}=440Hz*\lambda$

$\displaystyle \frac{340.29\frac{m}{s}}{440Hz}=\lambda$

$\displaystyle 0.773m=\lambda$

The relationship between velocity, frequency, and wavelength is:

$\displaystyle v=f\lambda$

Plug in the given information to solve:

$\displaystyle v=f\lambda$

$\displaystyle 340.29\frac{m}{s}=f*0.444m$

$\displaystyle \frac{340.29\frac{m}{s}}{0.444m}=f$

$\displaystyle 766.42Hz=f$

The relationship between frequency and period is $\displaystyle T=\frac{1}{f}$. 

Plug in our given value:

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

$\displaystyle T=\frac{1}{383Hz}$

$\displaystyle T=0.003s$

The relationship between velocity, frequency, and wavelength is:

$\displaystyle v=f\lambda$

In this case we're given a scenario where $\displaystyle v=f\lambda_1$ and $\displaystyle v=2f*\lambda_2$. The velocities equal each other because the problem states it has a constant velocity. Therefore we can set these equations equal to each other:

$\displaystyle f\lambda_1=2f\lambda_2$

Notice that the $\displaystyle f$'s cancel out:

$\displaystyle \lambda_1=2\lambda_2$

Divide both sides by two:

$\displaystyle \frac{1}{2}\lambda_1=\lambda_2$