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Chapter 15: Problem 10

A water wave traveling in a straight line on a lake is described by the equation $$y(x, t)=(3.75 \mathrm{cm}) \cos \left(0.450 \mathrm{cm}^{-1} x+5.40 \mathrm{s}^{-1} t\right)$$ where \(y\) is the displacement perpendicular to the undisturbed surface of the lake. (a) How much time does it take for one complete wave pattem 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

Expert verified
(a) Time: 1.163 s, Distance: 13.96 cm. (b) Wave number: 0.450 cm鈦宦�, Frequency: 0.86 Hz. (c) Wave speed: 12.00 cm/s, Floater max speed: 20.25 cm/s.

Step by step solution

01

Analyze the given wave equation

The wave equation is given by: \( y(x, t) = (3.75 \ \text{cm}) \cos(0.450 \ \text{cm}^{-1} x + 5.40 \ \text{s}^{-1} t) \). Here, the amplitude \( A = 3.75 \ \text{cm} \), wave number \( k = 0.450 \ \text{cm}^{-1} \), and angular frequency \( \omega = 5.40 \ \text{s}^{-1} \) are identified from the equation.
02

Calculate the period of the wave

The period \( T \) is the time it takes for one complete wave cycle to pass. It is calculated from the angular frequency using the formula \( T = \frac{2\pi}{\omega} \). Substitute \( \omega = 5.40 \ \text{s}^{-1} \) into the formula to get \( T = \frac{2\pi}{5.40} \approx 1.163 \ \text{s} \).
03

Calculate the wavelength of the wave

The wavelength \( \lambda \) is calculated from the wave number using the formula \( \lambda = \frac{2\pi}{k} \). Substitute \( k = 0.450 \ \text{cm}^{-1} \) to find \( \lambda = \frac{2\pi}{0.450} \approx 13.96 \ \text{cm} \).
04

Calculate the speed of the wave crest

The wave speed \( v \) can be calculated using the relation \( v = \frac{\lambda}{T} \), but it can also be directly calculated using \( v = \frac{\omega}{k} \). With \( \omega = 5.40 \ \text{s}^{-1} \) and \( k = 0.450 \ \text{cm}^{-1} \), \( v = \frac{5.40}{0.450} = 12.00 \ \text{cm/s} \).
05

Calculate the maximum speed of the cork floater

The maximum speed of the cork floater corresponds to the maximum speed of the vertical motion at the wave's peak. This is given by the formula \( v_{\text{max}} = A\omega \). Substitute \( A = 3.75 \ \text{cm} \) and \( \omega = 5.40 \ \text{s}^{-1} \) to get \( v_{\text{max}} = 3.75 \times 5.40 = 20.25 \ \text{cm/s} \).
06

Determine the number of waves per second

The frequency of the wave \( f \) is the number of complete waves passing a point per second, calculated by \( f = \frac{1}{T} \). From Step 2, \( T = 1.163 \ \text{s} \), so \( f \approx \frac{1}{1.163} \approx 0.86 \ \text{Hz} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Equation
Understanding wave motion starts with the wave equation, which describes how the wave travels. It conveys information about the wave's amplitude, wave number, and angular frequency. In the equation given as:
  • \( y(x, t) = (3.75 \, \text{cm}) \cos(0.450 \, \text{cm}^{-1} x + 5.40 \, \text{s}^{-1} t) \)
  • Ampitude \( A \) is 3.75 cm, which represents the maximum displacement of the wave from its rest position.
  • Wave number \( k \), here 0.450 cm\(^-1\), determines the number of oscillations per unit distance.
  • Angular frequency \( \omega \), which is 5.40 s\(^{-1}\), indicates how fast the wave oscillates in radians per second.
The cosine function, \( \cos \), in the wave equation signifies that the wave is oscillatory in nature, moving through phases as it propagates.
Angular Frequency
The angular frequency \( \omega \) is a vital component in understanding wave motion. It describes how many radians per second the wave oscillates. Specifically, angular frequency is related to the frequency of the wave and provides insight into the wave's speed and energy.
  • It is given by \( \omega = 5.40 \, \text{s}^{-1} \) in the provided wave equation.
To understand angular frequency fully:
  • It is connected to the period \( T \) through the formula \( \omega = \frac{2\pi}{T} \).
  • High angular frequencies indicate a fast oscillating wave.
Knowing \( \omega \) helps determine both the wave speed and timing, crucial for calculating when a wave reaches any given point.
Wave Speed
Wave speed connects the motion of the wave through space to time. It defines how quickly a wave crest travels from one point to another.
  • Wave speed \( v \) can be calculated using several methods:
    • \( v = \frac{\lambda}{T} \): where \( \lambda \) is the wavelength and \( T \) is the period.
    • \( v = \frac{\omega}{k} \): which provides a direct relation using angular frequency \( \omega \) and wave number \( k \).
  • In this scenario, the wave speed is determined as \( 12.00 \, \text{cm/s} \).
Understanding wave speed is essential for predicting how fast waves will move across a medium, such as water, and it reveals the dynamics of wave motion in real time.
Wavelength
Wavelength \( \lambda \) is a fundamental property of wave motion that describes the distance between successive crests of the wave or any two similar points.
  • Calculated from the wave number \( k \) using the formula \( \lambda = \frac{2\pi}{k} \).
  • Here, \( k = 0.450 \, \text{cm}^{-1} \), leading to a wavelength of approximately \( 13.96 \, \text{cm} \).
Wavelength is crucial because:
  • It helps determine the spatial layout of the wave.
  • Affects the wave speed when combined with the period.
Knowing the wavelength allows for understanding not just how far the wave travels per cycle, but also provides insight into the wave's energy and frequency.

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Most popular questions from this chapter

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