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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 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?

A \(1.50 \mathrm{~m}\) string of weight \(0.0125 \mathrm{~N}\) is tied to the ceiling at its upper end, and the lower end supports a weight \(W .\) Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$ y(x, t)=(8.50 \mathrm{~mm}) \cos (172 \mathrm{rad} / \mathrm{m} x-4830 \mathrm{rad} / \mathrm{s} t) $$ Assume that the tension of the string is constant and equal to \(W\). (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W ?\) (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling down the string?

A deep-sea diver is suspended beneath the surface of Loch Ness by a \(100-\mathrm{m}\) -long cable that is attached to a boat on the surface (Fig. \(\mathbf{P} 15.77\) ). The diver and his suit have a total mass of \(120 \mathrm{~kg}\) and a volume of \(0.0800 \mathrm{~m}^{3} .\) The cable has a diameter of \(2.00 \mathrm{~cm}\) and a linear mass density of \(\mu=1.10 \mathrm{~kg} / \mathrm{m} .\) The diver thinks he sees something moving in the murky depths and jerks the end of the cable back and forth to send transverse waves up the cable as a signal to his companions in the boat. (a) What is the tension in the cable at its lower end, where it is attached to the diver? Do not forget to include the buoyant force that the water (density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) ) exerts on him. (b) Calculate the tension in the cable a distance \(x\) above the diver. In your calculation, include the buoyant force on the cable. (c) The speed of transverse waves on the cable is given by \(v=\sqrt{F / \mu}\) [Eq. (15.14)]. The speed therefore varies along the cable, since the tension is not constant. (This expression ignores the damping force that the water exerts on the moving cable.) Integrate to find the time required for the first signal to reach the surface.

An ant with mass \(m\) is standing peacefully on top of a horizontal, stretched rope. The rope has mass per unit length \(\mu\) and is under tension \(F\). Without warning, Cousin Throckmorton starts a sinusoidal transverse wave of wavelength \(\lambda\) propagating along the rope. The motion of the rope is in a vertical plane. What minimum wave amplitude will make the ant become momentarily weightless? Assume that \(m\) is so small that the presence of the ant has no effect on the propagation of the wave.

The speed of sound in air at \(20^{\circ} \mathrm{C}\) is \(344 \mathrm{~m} / \mathrm{s}\). (a) What is the wavelength of a sound wave with a frequency of \(784 \mathrm{~Hz}\), corresponding to the note \(\mathrm{G}_{5}\) on a piano, and how many milliseconds does each vibration take? (b) What is the wavelength of a sound wave one octave higher (twice the frequency) than the note in part (a)?
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