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Chapter 12: Problem 4

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes \(2.5 \mathrm{~s}\) for the boat to travel from its highest point to its lowest, a total distance of \(0.62 \mathrm{~m}\). 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) Speed unchanged; amplitude is 0.15 m.

Step by step solution

01

Find the Wave Period

The boat takes \(2.5 \text{ s}\) to travel from the highest to the lowest point. Hence, the complete wave period \(T\) (one full cycle) is twice this time because it consists of both ascent and descent. Thus, \(T = 2 \times 2.5 \text{ s} = 5 \text{ s}\).
02

Calculate Wave Speed

The speed of a wave \(v\) is given by the formula \(v = \frac{\text{wavelength}}{\text{period}}\). We know the wavelength is given as \(6.0\text{ m}\) and the period \(T\) is \(5\text{ s}\). Therefore, the wave speed \(v = \frac{6.0\text{ m}}{5\text{ s}} = 1.2\text{ m/s}\).
03

Determine Wave Amplitude

Amplitude \(A\) is half the total vertical distance traveled by the boat, \(0.62 \text{ m}\). Therefore, the amplitude \(A = \frac{0.62 \text{ m}}{2} = 0.31 \text{ m}\).
04

Effect of Shorter Vertical Distance on Wave Speed

If the total vertical distance is \(0.30\text{ m}\), we first note that wave speed is dependent on wavelength and period only (not the vertical motion), so the speed remains \(1.2\text{ m/s}\).
05

Effect of Shorter Vertical Distance on Amplitude

With a total vertical travel of \(0.30 \text{ m}\), the amplitude is half of this distance. Thus, the new amplitude is \(A = \frac{0.30 \text{ m}}{2} = 0.15 \text{ m}\).

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Key Concepts

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

Wave Speed
Wave speed is a fundamental concept in wave mechanics, indicating how fast the wave crests move across the water surface. To calculate this, we use the formula:\[ v = \frac{\text{wavelength}}{\text{period}} \]
  • **Wavelength** refers to the distance between successive crests of a wave, here given as \(6.0 \text{ m}\).
  • **Period** is the time it takes for one full cycle, which involves both an ascent and descent. As solved in the steps, the period is \(5 \text{ s}\).
  • The resulting wave speed, using the given values, calculates to \(1.2 \text{ m/s}\).
This formula shows that wave speed depends on how far apart the crests are and how long it takes for the wave to complete a full cycle. Changes in these factors can adjust the wave speed significantly.
Wave Amplitude
Amplitude in wave mechanics measures the height of the wave from its average position to a crest, or from the average position to a trough. It's like the wave's 'tallness'. Given the boat's movement, the total vertical distance was \(0.62 \text{ m}\), and amplitude is half of this distance:\[ A = \frac{0.62 \text{ m}}{2} = 0.31 \text{ m} \]
  • An amplitude of \(0.31 \text{ m}\) is determined by dividing the full vertical travel distance by two.
  • This calculation implies that each crest and trough are \(0.31 \text{ m}\) from the calm water level.
If the vertical distance changes, as specified to \(0.30 \text{ m}\), the amplitude becomes \(0.15 \text{ m}\). Fluctuations in total distance directly adjust amplitude values since the amplitude is proportionate to vertical travel.
Wavelength
The wavelength is the spatial period of the wave鈥攖he distance over which the wave's shape repeats. For our fishing boat scenario, the fisherman observes the spacing between crests, measured as \(6.0 \text{ m}\). This constant measurement is pivotal in understanding how distant each successive wave is from another.
  • Having a wavelength of \(6.0 \text{ m}\) means each crest is \(6.0 \text{ m}\) apart from the next one.
  • This value aids in calculating wave speed, as it ties into the wave speed formula with the period.
Understanding wavelength helps unpack the physical properties of the wave, providing insight into the energy and dynamics present at the water's surface.
Wave Period
The wave period represents the time it takes to complete one full wave cycle, including an upward and downward movement for the boat in this situation. The exercise determines that traveling from top to bottom takes \(2.5 \text{ s}\), thus making the full period for one wave cycle:\[ T = 2 \times 2.5 \text{ s} = 5 \text{ s} \]
  • Completing a full cycle in \(5 \text{s}\) suggests that the wave continues this rhythm down the line.
  • The period is critical for precisely calculating the wave speed, combined with the known wavelength of \(6.0 \text{ m}\).
Thus, understanding wave period facilitates quantifying how quickly waves propagate and aids in predicting boat movement for safety and navigation purposes.

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Most popular questions from this chapter

Tuning a cello. A cellist tunes the C-string of her instrument to a fundamental frequency of \(65.4 \mathrm{~Hz}\). 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 she stretch that portion of the string? (b) What percentage 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 \(\mathrm{D} ?\)

A container ship is traveling westward at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). The waves on the surface of the ocean have a wavelength of \(40.0 \mathrm{~m}\) and are traveling eastward at a speed of \(16.5 \mathrm{~m} / \mathrm{s}\). (a) At what time intervals does the ship encounter the crest of a wave? (b) At what time intervals will the ship encounter wave crests if it turns around and heads eastward?

Two train whistles, \(A\) and \(B\), each have a frequency of \(392 \mathrm{~Hz}\). \(A\) is stationary and \(B\) is moving toward the right (away from \(A\) ) at a speed of \(35.0 \mathrm{~m} / \mathrm{s}\). A listener is between the two whistles and is moving toward the right with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). (See Figure \(12.43 .)\) (a) What is the frequency from \(A\) as heard by the listener? (b) What is the frequency from \(B\) as heard by the listener? (c) What is the beat frequency detected by the listener?

The longest pipe found in most medium-sized pipe organs is \(4.88 \mathrm{~m}\) (16 ft). What is the wavelength of the note corresponding to the fundamental mode if the pipe is (a) open at both ends and (b) open at one end and closed at the other?

While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an identical horn, approaching you. You blow your horn, which has a frequency of \(260 \mathrm{~Hz}\); your friend begins to blow his horn as well, and you hear a beat frequency of \(6.0 \mathrm{~Hz}\). How fast is your friend approaching you?
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