Question

A water wave traveling in a straight line on a lake is described by the equation $$y(x, t) = (\mathrm{2.75 \, cm) cos(0.410 \mathrm{rad/cm} \, \textit{x}} + 6.20 \, \mathrm{rad}/s \; t)$$ where y is the displacement perpendicular to the undisturbed surface of the lake. (a) How much time does it take for one complete wave pattern to go past a fisherman in a boat at anchor, and what horizontal distance does the wave crest travel in that time? (b) What are the wave number and the number of waves per second that pass the fisherman? (c) How fast does a wave crest travel past the fisherman, and what is the maximum speed of his cork floater as the wave causes it to bob up and down?

Short answer

(a) Time is 1.013 s, distance is 15.32 cm. (b) Wave number is 0.410 rad/cm, frequency is 0.987 Hz. (c) Wave speed is 15.12 cm/s, max floater speed is 17.05 cm/s.

Step-by-step solution

Understanding the General Wave Equation

The wave is described by the equation \( y(x, t) = 2.75 \cos(0.410x + 6.20t) \). Here, the angular wave number \( k = 0.410 \, \text{rad/cm} \) and the angular frequency \( \omega = 6.20 \, \text{rad/s} \). The amplitude is \( 2.75 \, \text{cm} \).

Calculate the Time Period of the Wave

The time period \( T \) of the wave is the time taken for one complete cycle and is calculated using \( \omega = \frac{2\pi}{T} \). Rearranging gives \( T = \frac{2\pi}{\omega} = \frac{2\pi}{6.20} \).

Compute Horizontal Distance Traveled by Wave Crest

The wavelength \( \lambda \) can be found from the wave number using \( k = \frac{2\pi}{\lambda} \), so \( \lambda = \frac{2\pi}{0.410} \). The horizontal distance a wave crest travels in one period is exactly one wavelength.

Determine the Wave Number and Frequency

The wave number \( k = 0.410 \, \text{rad/cm} \) measures how many radians fit into one centimeter. The frequency \( f \) is \( f = \frac{1}{T} \).

Calculate Wave Speed

The speed of the wave crest, \( v \), can be found using \( v = f\lambda = \frac{\omega}{k} \).

Find Maximum Speed of Cork Floater

The vertical speed of the floater has its maximum value when \( \cos(0.410x + 6.20t) \) is zero, which gives maximum velocity \( v_{\text{max}} = \omega A = 6.20 \times 2.75 \).

Substitute Values to Complete Calculations

Perform the calculations using values: \( T = \frac{2\pi}{6.20} \), \( \lambda = \frac{2\pi}{0.410} \), \( f = \frac{1}{T} \), \( v = \frac{\omega}{k} \), and \( v_{\text{max}} = 6.20 \times 2.75 \). Results are \( T \approx 1.013 \, \text{s}, \lambda \approx 15.32 \, \text{cm}, f \approx 0.987 \, \text{Hz}, v \approx 15.12 \, \text{cm/s}, v_{\text{max}} \approx 17.05 \, \text{cm/s} \).

Key concepts

Angular Frequency

In wave motion, the angular frequency is a crucial concept. It helps us understand how fast the wave oscillates. Angular frequency, denoted by \( \omega \), is measured in radians per second (rad/s). It tells us how many radians the wave front notches up in one second.

For the given wave equation \( y(x, t) = 2.75 \cos(0.410x + 6.20t) \), the angular frequency \( \omega \) is provided as \( 6.20 \; \text{rad/s} \). This value tells us that each point on the wave completes 6.20 radians in a single second. It's important because it directly relates to the speed of wave crest movement across a point.

To calculate the time period \( T \) (how long it takes for one complete oscillation), we use the formula \( \omega = \frac{2\pi}{T} \). Solving for \( T \) gives us \( T = \frac{2\pi}{6.20} \), which is approximately 1.013 seconds.

Wave Number

The wave number, denoted by \( k \), is another important characteristic of wave motion. It represents the number of waves present in a unit distance. Measured in radians per centimeter (rad/cm), the wave number provides insight into how frequently the wave cycles occur over a given length.

In the wave equation provided, \( k = 0.410 \, \text{rad/cm} \). This tells us that for each centimeter of wave travel, the wave completes 0.410 radians. It's a measure of the wave's spatial frequency.

Additionally, the wave number is related to the wavelength through the equation \( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength. By rearranging, we find \( \lambda = \frac{2\pi}{0.410} \), resulting in approximately 15.32 cm for the wavelength.

Wavelength

Wavelength, denoted by \( \lambda \), is the distance between two consecutive points that are in phase on the wave, like crest to crest or trough to trough. This physical dimension of the wave gives us an understanding of the "size" of the wave's oscillation.

From our earlier calculation, we found that the wavelength \( \lambda = \frac{2\pi}{0.410} \) is approximately 15.32 cm. This indicates that every 15.32 cm, the wave completes one full cycle.

Understanding the wavelength is fundamental to knowing how the wave will interact with objects and obstacles it encounters. It also helps us calculate the wave's velocity by using the relation \( v = f\lambda \), where \( v \) is the velocity and \( f \) is the frequency.

Velocity of Waves

The velocity of a wave, \( v \), is the speed at which the wave crest moves over time. Wave velocity is calculated using the formula \( v = f\lambda \) or equivalently \( v = \frac{\omega}{k} \), linking the relationship between wave angular frequency, wave number, and the speed.

In our specific scenario, using the given \( \omega = 6.20 \, \text{rad/s} \) and \( k = 0.410 \, \text{rad/cm} \), we calculate the velocity as \( v = \frac{6.20}{0.410} \), which yields approximately 15.12 cm/s. This tells us how fast the wave crest travels past a stationary point, like our fisherman in the problem.

A crucial pointer: this velocity is constant for waves traveling through a uniform medium and remains unaffected by the wave's amplitude. Understanding wave velocity is important in predicting the wave's impact and interaction with its environment.