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Chapter 16: Problem 4

Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticeable, but upon reaching shore, they can rise up to the astonishing height of a six-story building. One tsunami, generated off the Aleutian islands in Alaska, had a wavelength of \(750 \mathrm{km}\) and traveled a distance of \(3700 \mathrm{km}\) in 5.3 h. (a) What was the speed (in \(\mathrm{m} / \mathrm{s}\) ) of the wave? For reference, the speed of a 747 jetliner is about \(250 \mathrm{m} / \mathrm{s} .\) Find the wave's (b) frequency and (c) period.

Short Answer

Expert verified
Wave speed 鈮� 193.92 m/s, frequency 鈮� 0.00025856 Hz, period 鈮� 3868.5 s.

Step by step solution

01

Convert Distance and Time to Compatible Units

First, convert the distance traveled by the tsunami from kilometers to meters. Also, convert the time from hours to seconds.1. Distance: \[ 3700 \, \text{km} = 3700 \, \text{km} \times 1000 \, \text{m/km} = 3,700,000 \, \text{m} \]2. Time: \[ 5.3 \, \text{hours} = 5.3 \, \text{hours} \times 3600 \, \text{s/hour} = 19,080 \, \text{s} \]
02

Calculate the Speed of the Wave

Now, use the formula for speed, which is the distance traveled divided by the time taken.\[\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{3,700,000 \, \text{m}}{19,080 \, \text{s}} \approx 193.92 \, \text{m/s}\]
03

Calculate the Frequency of the Wave

Frequency is calculated using the wave speed and the wavelength. The formula for frequency \( f \) is given by \( f = \frac{v}{\lambda} \), where \( v \) is the speed and \( \lambda \) is the wavelength.1. Convert wavelength from km to meters: \[ 750 \, \text{km} = 750 \, \text{km} \times 1000 \, \text{m/km} = 750,000 \, \text{m} \]2. Calculate the frequency: \[ f = \frac{193.92 \, \text{m/s}}{750,000 \, \text{m}} \approx 0.00025856 \, \text{Hz} \]
04

Calculate the Period of the Wave

The period \( T \) is the reciprocal of the frequency. Use the formula \( T = \frac{1}{f} \).\[T = \frac{1}{0.00025856 \, \text{Hz}} \approx 3868.5 \, \text{s}\]

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Key Concepts

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

Tsunami Speed
When it comes to understanding tsunami mechanics, speed is a critical aspect to discuss. Tsunamis are known for their rapid movement, traversing vast distances in a relatively short time. To calculate the speed of a tsunami, one needs to understand the basic speed formula:
  • Speed is calculated by the formula: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \).
  • For our exercise, having converted the distance (3,700 km to 3,700,000 meters) and time (5.3 hours to 19,080 seconds), the speed of the tsunami was found to be approximately 193.92 meters per second.
This speed is quite impressive, especially when compared to commercial aircraft, such as the 747 jetliner, which flies at around 250 m/s. Tsunamis in deep oceans move with such swiftness due to limited friction. However, as they approach shallower waters near shorelines, their speed decreases while their height increases dramatically, posing a huge potential hazard to coastal areas.
Wave Frequency
Frequency is a fundamental concept in wave mechanics, describing how many wave crests pass a fixed point per unit of time. To understand wave frequency:
  • Frequency is calculated using the formula: \( f = \frac{v}{\lambda} \), where \( v \) is wave speed and \( \lambda \) is wavelength.
  • In the context of our tsunami, where the speed is approximately 193.92 m/s and the wavelength is 750,000 meters (converted from 750 km), the frequency is about 0.00025856 Hz.
This frequency indicates a very long time between successive wave fronts in deep water, fitting the nature of tsunamis which, though infrequent in wave fronts, have significant energy and potency as they reach shore. Their destructive capability doesn't rely on high frequency but rather on energy spread over vast areas.
Wave Period
Another key wave feature is the period, which is the time taken for one complete cycle of the wave to pass a point. The wave period can be found with ease using the frequency:
  • The formula for Period is: \( T = \frac{1}{f} \).
  • For our tsunami, with a frequency of 0.00025856 Hz, the period calculates to about 3868.5 seconds.
This longer period is characteristic of tsunamis, as they are often described as having a 'long wavelength' and 'long period.' What this means practically is that, although tsunamis travel quickly, the amount of time it takes for an entire wave cycle to occur is quite long鈥攐ver an hour in this case. As a result, it provides a deceptive calm between waves of destruction, giving little warning for another powerful wave to hit after the first.

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Most popular questions from this chapter

Have you ever listened for an approaching train by kneeling next to a railroad track and putting your ear to the rail? Young's modulus for steel is \(Y=2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2},\) and the density of steel is \(\rho=7860 \mathrm{kg} / \mathrm{m}^{3} .\) On a day when the temperature is \(20^{\circ} \mathrm{C},\) how many times greater is the speed of sound in the rail than in the air?

In a discussion person A is talking 1.5 dB louder than person B, and person \(\mathrm{C}\) is talking \(2.7 \mathrm{dB}\) louder than person A. What is the ratio of the sound intensity of person \(\mathrm{C}\) to the sound intensity of person \(\mathrm{B} ?\)

A bat emits a sound whose frequency is \(91 \mathrm{kHz}\). The speed of sound in air at \(20.0^{\circ} \mathrm{C}\) is \(343 \mathrm{m} / \mathrm{s} .\) However, the air temperature is \(35^{\circ} \mathrm{C},\) so the speed of sound is not \(343 \mathrm{m} / \mathrm{s}\). Assume that air behaves like an ideal gas, and find the wavelength of the sound.

A spider hangs from a strand of silk whose radius is \(4.0 \times 10^{-6} \mathrm{m}\). The density of the silk is \(1300 \mathrm{kg} / \mathrm{m}^{3} .\) When the spider moves, waves travel along the strand of silk at a speed of \(280 \mathrm{m} / \mathrm{s}\). Ignore the mass of the silk strand, and determine the mass of the spider.

Using an intensity of \(1 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\) as a reference, the threshold of hearing for an average young person is 0 dB. Person 1 and person 2 , who are not average, have thresholds of hearing that are \(\beta_{1}=-8.00 \mathrm{dB}\) and \(\beta_{2}=+12.0 \mathrm{dB}\). What is the ratio \(I_{1} / I_{2}\) of the sound intensity \(I_{1}\) when person 1 hears the sound at his own threshold of hearing compared to the sound intensity \(I_{2}\) when person 2 hears the sound at his own threshold of hearing?
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