Problem 1
Question
Suppose you are standing on the end of a pier watching the waves and, between your position and a buoy \(200 \mathrm{~m}\) straight out, you count 15 wave crests. Further, suppose a wave crest comes by every 15 seconds. Calculate \(\nu\) in \(\mathrm{Hz}, \lambda\) in \(\mathrm{m}, c\) in \(\mathrm{m} \mathrm{sec}^{-1}\), and \(\bar{\nu}\) in \(\mathrm{km}^{-1}\).
Step-by-Step Solution
Verified Answer
\(\nu \approx 0.067 \text{ Hz}, \lambda = 13.33 \text{ m}, c \approx 0.89 \text{ m/s}, \bar{\nu} = 75 \text{ km}^{-1}.\)
1Step 1: Calculate the Wavelength (λ)
The distance across which the wave crests are counted is 200 meters, with 15 wave crests in this distance. The distance covered by one complete wave crest to the next (wavelength) is given by dividing the total distance by the number of waves: \[ \lambda = \frac{200}{15} = 13.33 \text{ meters} \]
2Step 2: Calculate the Frequency (ν)
Frequency is defined as the number of wave crests passing a point per second. Since a wave crest passes every 15 seconds, the frequency is the inverse of this period: \[ u = \frac{1}{15} \text{ Hz} \approx 0.067 \text{ Hz} \]
3Step 3: Calculate the Wave Speed (c)
Wave speed is obtained from the product of its frequency and wavelength. Using our values for \(u\) and \(\lambda\): \[ c = u \cdot \lambda = 0.067 \times 13.33 \approx 0.89 \text{ m/sec} \]
4Step 4: Calculate Wave Number (\(\bar{\nu}\))
Wavenumber is defined as the number of wavelengths per meter, or the inverse of the wavelength, converted to kilometers to match the desired unit. \[ \bar{u} = \frac{1}{\lambda} = \frac{1}{13.33} \approx 0.075 \text{ m}^{-1} = 75 \text{ km}^{-1} \]
5Step 5: Convert Wavenumber to Required Unit
Since the given problem requests \(\bar{u}\) in \(\text{km}^{-1}\), there isn't a need for further conversion from the previous step.
Key Concepts
WavelengthFrequencyWave SpeedWavenumber
Wavelength
The wavelength of a wave is a fundamental concept in wave physics. It represents the distance between successive points of a wave that are in phase; for example, the distance between two successive crests. In our exercise, this distance is calculated as the wavelength (\( \lambda \)). By counting 15 wave crests over a 200-meter stretch and dividing this distance by the number of waves, you obtain:
- \( \lambda = \frac{200}{15} \approx 13.33 \text{ meters} \)
Frequency
Frequency relates to how often the waves are passing a certain point over time and is crucial for understanding wave behavior. It's expressed in Hertz (Hz), where 1 Hz equals 1 cycle per second. Frequency (\( u \)) can easily be found as the inverse of the wave period - the time taken for one full wave cycle to pass a point. In this scenario, since a wave crest comes by every 15 seconds, the frequency is:
- \( u = \frac{1}{15} \text{ Hz} \approx 0.067 \text{ Hz} \)
Wave Speed
The speed of a wave is determined by how quickly the wave travels through the medium. It combines both the frequency and wavelength to express this movement's velocity. Mathematically, wave speed (\( c \)) is calculated using the formula:
- \( c = u \cdot \lambda \)
- \( c = 0.067 \times 13.33 \approx 0.89 \text{ m/sec} \)
Wavenumber
The concept of wavenumber is essential for identifying how many wavelengths fit into a given unit of length. It is essentially the spatial frequency of the wave, representing how many cycles occur over a unit distance. In our context, the wavenumber (\( \bar{u} \)) is found by taking the inverse of the wavelength and converting it when necessary:
- \( \bar{u} = \frac{1}{\lambda} \)
- \( \bar{u} = \frac{1}{13.33} \approx 0.075 \text{ m}^{-1} = 75 \text{ km}^{-1} \)
Other exercises in this chapter
Problem 2
Blue light has \(\bar{\nu}=20,800 \mathrm{~cm}^{-1}\). Calculate \(\nu\) in \(\mathrm{Hz}\) and \(\lambda\) in \(\mathrm{nm}\).
View solution Problem 3
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