Problem 11 - Physics

Question

A water-skier is moving at a speed of \(12.0 \mathrm{~m} / \mathrm{s}\). When she skis in the same direction as a traveling wave, she springs upward every \(0.600 \mathrm{~s}\) because of the wave crests. When she skis in the direction opposite to that in which the wave moves, she springs upward every \(0.500 \mathrm{~s}\) in response to the crests. The speed of the skier is greater than the speed of the wave. Determine (a) the speed and (b) the wavelength of the wave.

Step-by-Step Solution

Answer

The wave speed is 8 m/s and the wavelength is 4 m.

Step 1: Define the problem

We need to determine two things: the speed of the wave and its wavelength. The skier's upward movement depends on the frequency of encountering wave crests, which changes based on her direction relative to the wave.

Step 2: Establish the formula for frequency observed when moving

The formula to find the observed frequency when moving towards or away from a wave is: \(f = \frac{v_{wave}}{\lambda} \pm \frac{v_{skier}}{\lambda}\), where the positive sign is used when moving against the wave and the negative when moving with the wave.

Step 3: Set up equations for frequencies

When moving in the direction of the wave, the frequency observed is \(\frac{1}{0.6}\) Hz (as the skier hits a crest every 0.6 seconds) and against the wave is \(\frac{1}{0.5}\) Hz.

Step 4: Solve for speed of the wave

Using the equations from Step 3, we have:\[\frac{1}{0.6} = \frac{v_{wave} - 12}{\lambda} \quad \text{and} \quad \frac{1}{0.5} = \frac{v_{wave} + 12}{\lambda}.\]Rearrange each equation to equal \(v_{wave}\) and then set them equal to solve for \(\lambda\).

Step 5: Calculate the speed of the wave

Subtract the first equation from the second to eliminate \(\lambda\):\[\frac{1}{0.5} - \frac{1}{0.6} = \frac{(v_{wave} + 12) - (v_{wave} - 12)}{\lambda}.\]Simplify and solve for \(v_{wave}\):\[\frac{6}{3} = \frac{24}{\lambda} \Rightarrow \lambda = 4 \, m.\]Using one of the original frequency equations:\[\frac{1}{0.5} = \frac{v_{wave} + 12}{4}.\]Solve for \(v_{wave}:\) \[v_{wave} = 8 \, \text{m/s}.\]

Step 6: Determine the wavelength

We've already found the wavelength is 4 meters from our elimination process in Step 5.

Key Concepts

Frequency

The frequency of a wave is the number of times a wave crest passes a given point in one second. It is measured in Hertz (Hz).
For the water-skier example, the frequency changes depending on her direction relative to the wave.

• When skiing in the same direction as the wave, the skier encounters the crests less frequently, resulting in a lower frequency.
• Conversely, when skiing against the wave, the crests reach her more frequently, which increases the frequency.

The frequencies observed were \(\frac{1}{0.6}\) Hz and \(\frac{1}{0.5}\) Hz, respectively. These frequencies allow us to understand how quickly the skier is interacting with the wave crests based on her direction of movement.

Wavelength

Wavelength is the distance between two consecutive points in phase on a wave, such as from crest to crest or trough to trough. It is usually denoted by the symbol \(\lambda\) and measured in meters.
In the water-skier problem, calculating the wavelength involves using the skier's speed and the difference in observed frequency.

• When the skier is skiing in the direction of the wave, she takes longer to meet the wave crests.
• When skiing against the direction, she meets them more often.

The equation \(\frac{v_{wave} - 12}{\lambda} \) for skiing with the wave and \(\frac{v_{wave} + 12}{\lambda} \) for skiing against it helps calculate the wavelength, giving us \(\lambda = 4 \, \text{m}\), indicating how long the wave crests are from each other.

Doppler Effect

The Doppler Effect explains how the frequency of a wave changes relative to an observer moving towards or away from the wave source.
For the water-skier, as she moves, the frequency of the encountered wave crests changes based on her speed relative to the wave.

• Moving towards the wave source (or the wave moving towards the observer) increases frequency because the waves are 'bunched up.'
• Moving away decreases frequency since the waves are 'spread out.'

This principle is used to find the apparent frequency from the equations \(f = \frac{v_{wave}}{\lambda} \pm \frac{v_{skier}}{\lambda} \), demonstrating the practical effect of the Doppler shift in wave phenomenon.

Wave Speed

Wave speed is the rate at which a wave travels through a medium. It is calculated using the relationship between frequency and wavelength.
In the skier's case, wave speed \(v_{wave}\) can be found using formulas:

• When moving with the wave, \(\frac{1}{0.6} = \frac{v_{wave} - 12}{\lambda} \)
• When moving against the wave, \(\frac{1}{0.5} = \frac{v_{wave} + 12}{\lambda} \)

By solving these equations, we determined \(v_{wave} = 8 \, \text{m/s}\). This speed tells us how fast the wave travels on the water, independent of the skier's motion. Understanding wave speed is crucial for predicting wave behavior in various environments.