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Chapter 15: 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 \(0.53 \mathrm{~m} .\) The fisherman sees that the wave crests are spaced \(4.8 \mathrm{~m}\) 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) change?

Short Answer

Expert verified
The speed of the waves is \(0.96m/s\), the amplitude of each wave is \(0.265m\), and if the total vertical distance travelled by the boat were \(0.30m\), the speed would remain \(0.96m/s\) while the amplitude would be \(0.15m\).

Step by step solution

01

Calculate the wave speed

The speed of the waves can be found using the formula: Speed = Wavelength / Period. The wavelength is the distance between wave crests, which is given as 4.8m. The period is twice the time it takes for the boat to travel from its highest to its lowest point, so the period is \(2*2.5 = 5.0s\). So the speed of the waves is \(4.8m / 5.0s = 0.96m/s\).
02

Calculate the amplitude of each wave

The amplitude is the maximum displacement from equilibrium, which is half the total up-and-down distance travelled by the boat. This is given as 0.53m, so the amplitude of each wave is \(0.53m / 2 = 0.265m\).
03

Calculate the altered wave speed and amplitude

If the total vertical distance travelled by the boat were 0.30m and the other data remained the same, the period and wavelength(and therefore the speed) would not change, as these are independent of the motion of the boat. Hence, the speed remains \(0.96m/s\). However, the amplitude would be half of the new vertical distance, so the new amplitude is \(0.30m / 2 = 0.15m\).

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Key Concepts

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

Wave Speed
Understanding wave speed is pivotal when studying wave physics. It is defined as the distance a wave covers over a unit of time.

In our example, the wave speed calculation was based on the relationship between wavelength and wave period. The formula is given by: \[ \text{Speed} = \frac{\text{Wavelength}}{\text{Period}} \.\] The fisherman observes that the distance between successive wave crests (wavelength) is 4.8 meters, and the time for one full cycle (period) is 5 seconds, leading us to a wave speed of 0.96 meters per second (m/s). This simple calculation can be applied universally to understand how fast waves in any medium are traveling.
Wave Amplitude
Wave amplitude is essentially a measure of the wave's energy. In terms of a water wave, it's the height from the water's rest position to the crest of the wave.

To find the amplitude, we consider the total vertical movement of the boat, which is 0.53 meters. Since the boat moves from the highest point (crest) to the lowest (trough), the amplitude is half of this distance. Using the formula \[ \text{Amplitude} = \frac{\text{Vertical Distance}}{2} \.\] we find that the amplitude of the wave is 0.265 meters. This reflects how far the particles within the medium (in this case, water) are displaced due to the wave's energy.
Wave Period
The wave period is the amount of time it takes for a full cycle of the wave to pass a certain point. We can think of it as the 'heartbeat' of the wave, as it reflects the regularity of the wave cycles.

In the exercise, the period was calculated by doubling the time observed for the boat to move from the crest to the trough, resulting in a period of 5 seconds. This interval is crucial as it determines how often the wave oscillations occur and is intrinsic to calculating the wave speed and energy.
Wave Wavelength
Last but not least, the wave wavelength is the spatial period of the wave: the distance over which the wave's shape repeats. It is measured from any point on one wave to the corresponding point on the adjacent wave, such as from crest to crest or trough to trough.

From the fisherman's observations, the wavelength is discerned to be 4.8 meters. Knowing the wavelength and the period, we can understand how wave speed is determined and also predict where the wave will be after a certain time interval, which is essential for many applications in wave physics.

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Most popular questions from this chapter

A horizontal wire is tied to supports at each end and vibrates in its second- overtone standing wave. The tension in the wire is \(5.00 \mathrm{~N}\), and the node-to-node distance in the standing wave is \(6.28 \mathrm{~cm}\). (a) What is the length of the wire? (b) A point at an antinode of the standing wave on the wire travels from its maximum upward displacement to its maximum downward displacement in \(8.40 \mathrm{~ms}\). What is the wire's mass?

A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is \(0.386 \mathrm{~m}\). The maximum transverse acceleration of a point at the middle of the segment is \(8.40 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\) and the maximum transverse velocity is \(3.80 \mathrm{~m} / \mathrm{s}\). (a) What is the amplitude of this standing wave? (b) What is the wave speed for the transverse traveling waves on this string?

Adjacent antinodes of a standing wave on a string are \(15.0 \mathrm{~cm}\) apart. A particle at an antinode oscillates in simple harmonic motion with amplitude \(0.850 \mathrm{~cm}\) and period \(0.0750 \mathrm{~s}\). The string lies along the \(+x\) -axis and is fixed at \(x=0 .\) (a) How far apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern? (c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

Transverse waves on a string have wave speed \(8.00 \mathrm{~m} / \mathrm{s},\) amplitude \(0.0700 \mathrm{~m},\) and wavelength \(0.320 \mathrm{~m} .\) The waves travel in the \(-x\) -direction, and at \(t=0\) the \(x=0\) end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves. (b) Write a wave function describing the wave. (c) Find the transverse displacement of a particle at \(x=0.360 \mathrm{~m}\) at time \(t=0.150 \mathrm{~s}\). (d) How much time must elapse from the instant in part (c) until the particle at \(x=0.360 \mathrm{~m}\) next has maximum upward displacement?

The wave function of a standing wave is \(y(x, t)=(4.44 \mathrm{~mm})\) \(\sin [(32.5 \mathrm{rad} / \mathrm{m}) x] \sin [(754 \mathrm{rad} / \mathrm{s}) t] .\) For the two traveling waves that make up this standing wave, find the (a) amplitude; (b) wavelength; (c) frequency; (d) wave speed; (e) wave functions. (f) From the information given, can you determine which harmonic this is? Explain.
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