/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A fisherman notices that his boa... [FREE SOLUTION] | 91Ó°ÊÓ

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A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s for the boat to travel from its highest point to its lowest, a total distance of 0.62 m. The fisherman sees that the wave crests are spaced 6.0 \(\mathrm{m}\) apart. (a) How fast are the waves traveling? (b) What is the amplitude of each wave? (c) If the total vertical distance traveled by the boat were 0.30 \(\mathrm{m}\) but the other data remained the same, how would the answers to parts (a) and (b) be affected?

Short Answer

Expert verified
(a) 1.2 m/s, (b) 0.31 m, (c) Wave speed remains 1.2 m/s, amplitude is 0.15 m.

Step by step solution

01

Understand the general information about waves

The wave motion can be analyzed in terms of wavelength, frequency, speed, and amplitude. Here, we have the time it takes to move from the highest to lowest point (half a period), the distance between wave crests (wavelength), and the total vertical distance the boat moves (twice the amplitude).
02

Calculate the wave speed (Part a)

The wave speed can be calculated using the formula: \[ v = \frac{\lambda}{T} \]where \( \lambda \) is the wavelength and \( T \) is the period (full time for an up-and-down cycle). We know the time from highest to lowest is half the period, so: \[ T = 2 \times 2.5s = 5s \] Given, \( \lambda = 6.0m \), then:\[ v = \frac{6.0m}{5s} = 1.2 \text{ m/s} \]
03

Determine the amplitude (Part b)

Amplitude is half of the total vertical distance traveled by the boat. Given that the total distance is 0.62 m, the amplitude is:\[ A = \frac{0.62m}{2} = 0.31m \]
04

Re-evaluate conditions with new vertical distance (Part c)

If the total vertical distance is 0.30 m, the amplitude becomes:\[ A' = \frac{0.30m}{2} = 0.15m \]The wave speed remains unaffected because it depends only on the wavelength and the period, both of which are unchanged. Thus, \[ v = 1.2 \text{ m/s}\] remains the same.

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Key Concepts

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

Wavelength
When we talk about waves, one fundamental concept is the **wavelength**. This refers to the distance between consecutive crests (or troughs) in a wave. In simpler terms, it's the length of one full wave cycle. Imagine waves on a beach, the distance from one crest to the next is the wavelength. In the fisherman's situation, he notes that the wave crests are 6.0 meters apart. This measurement, 6.0 meters, is therefore the wavelength of the wave.
  • Wavelength is denoted by the symbol \( \lambda \) (lambda).
  • It is usually measured in meters.
Understanding wavelength helps us relate it to other important properties of waves, like their speed and frequency.
Wave Speed
**Wave speed** is essentially how fast a wave travels through a medium. It's like the speed of a car on a road—in this case, the water's surface is the road, and the wave is the car. To calculate wave speed, we need to know both the wavelength and the period of the wave, which is the time it takes for one full cycle of motion.
We use the formula: \[ v = \frac{\lambda}{T} \] where \( v \) is the wave speed, \( \lambda \) is the wavelength, and \( T \) is the period.
In the exercise, given that the boat takes 5 seconds for a full cycle and the wavelength is 6.0 meters, we can calculate the wave speed as:
  • \( v = \frac{6.0 \, \text{m}}{5 \, \text{s}} = 1.2 \, \text{m/s} \)
Wave speed tells us how quickly the wave crests reach the fisherman, which in this problem, is 1.2 meters per second, indicating that the waves are rolling towards him at this speed.
Amplitude
The **amplitude** of a wave is essential in describing its intensity. It measures how far the wave moves from its central rest position, reaching up to the crest or down to the trough. In our scenario, the amplitude of the wave corresponds to how high the boat rises or falls due to the waves.
This problem gives us the total vertical distance (from top to bottom), which is 0.62 meters. Since amplitude represents half of this vertical distance, we calculate amplitude as:
  • \( A = \frac{0.62 \, \text{m}}{2} = 0.31 \, \text{m} \)
Changes in amplitude affect how much the boat moves up and down, but not the wave speed. If the total distance changes to 0.30 meters, the new amplitude would be:
  • \( A' = \frac{0.30 \, \text{m}}{2} = 0.15 \, \text{m} \)
Amplitude, therefore, represents the drama of the wave without altering its speed.
Frequency
The **frequency** of a wave is an exciting concept that tells us about the number of wave cycles that occur in one second. It’s like counting how many waves hit the shore in a minute if you’re sitting on the beach. Frequency is closely related to the wave speed and wavelength through the formula:
\[ f = \frac{1}{T} \]
Here, \( f \) is the frequency and \( T \) is the period, or the time for one full wave cycle. Since we derived the period as 5 seconds earlier, the frequency would be:
  • \( f = \frac{1}{5 \, \text{s}} = 0.2 \, \text{Hz} \)
Frequency is measured in Hertz (Hz), where one Hertz is equivalent to one cycle per second. Therefore, a frequency of 0.2 Hz means that there’s almost one wave cycle every 5 seconds.
Understanding frequency provides insight into the rhythmic nature of wave motion, which adds to the understanding of phenomena like sound and light waves, as well as the water waves the fisherman experiences.

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Most popular questions from this chapter

Tuning an Instrument. A musician tunes the C-string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 \(\mathrm{m}\) long and has a mass of 14.4 \(\mathrm{g} .\) (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 \(\mathrm{Hz}\) to 73.4 \(\mathrm{Hz}\) , corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D} ?\)

CALC A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t)=\) \((5.60 \mathrm{cm}) \sin [(0.0340\) rad/ \(\mathrm{cm}) x] \sin [(50.0 \mathrm{rad} / \mathrm{s}) t],\) where the origin is at the left end of the string, the \(x\) -axis is along the string. and the \(y\) -axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x, t)\) for this string if it were vibrating in its eighth harmonic?

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