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Chapter 15: Problem 18

(II) The intensity of an earthquake wave passing through the Earth is measured to be \(3.0 \times 10^{6} \mathrm{~J} / \mathrm{m}^{2} \cdot \mathrm{s}\) at a distance of \(48 \mathrm{~km}\) from the source. (a) What was its intensity when it passed a point only \(1.0 \mathrm{~km}\) from the source? \((b)\) At what rate did energy pass through an area of \(2.0 \mathrm{~m}^{2}\) at \(1.0 \mathrm{~km} ?\)

Short Answer

Expert verified
(a) \(6.912 \times 10^9 \mathrm{~J/m^2\cdot s}\); (b) \(1.3824 \times 10^{10} \mathrm{~J/s}\).

Step by step solution

01

Understand the Problem

We're given the intensity of an earthquake wave at a certain distance and asked to find the intensity at a closer distance. Additionally, we need to calculate the energy rate through a given area at the closer distance.
02

Use the Inverse Square Law

The intensity of a wave diminishes with distance according to the inverse square law: \( I_2 = I_1 \left(\frac{r_1}{r_2}\right)^2 \). Here, \( I_1 = 3.0 \times 10^6 \mathrm{~J} / \mathrm{m}^2 \cdot \mathrm{s} \), \( r_1 = 48 \mathrm{~km} \), and \( r_2 = 1.0 \mathrm{~km} \).
03

Calculate the Intensity at 1 km

Substitute \( I_1 \), \( r_1 \), and \( r_2 \) into the inverse square law formula: \[ I_2 = 3.0 \times 10^6 \left(\frac{48}{1}\right)^2 = 3.0 \times 10^6 \times 2304 = 6.912 \times 10^9 \mathrm{~J} / \mathrm{m}^2 \cdot \mathrm{s} \].
04

Calculate the Energy Rate through the Area

Once we have \( I_2 = 6.912 \times 10^9 \mathrm{~J} / \mathrm{m}^2 \cdot \mathrm{s} \), we can find the energy rate through the area of \(2.0 \mathrm{~m}^2\) using the formula: \( P = I \times A \). Substitute the intensity and area: \[ P = 6.912 \times 10^9 \times 2.0 = 1.3824 \times 10^{10} \mathrm{~J/s} \].
05

Final Step: Verify Calculations

Ensure all calculations follow correctly from the inverse square law and multiplication to avoid significant errors.

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Key Concepts

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

Inverse Square Law
The Inverse Square Law is a fundamental principle that explains how the intensity of a wave decreases as the distance from the source increases. This principle applies to various types of waves, including earthquake waves, sound waves, and light waves. According to the law:
\[I_2 = I_1 \left(\frac{r_1}{r_2}\right)^2\]where:
  • \( I_1 \) is the initial intensity of the wave at a distance \( r_1 \)
  • \( I_2 \) is the intensity at the distance \( r_2 \)
  • \( r_1 \) and \( r_2 \) are the initial and new distances from the source, respectively
This mathematical relationship shows that as you move farther away from the source, the intensity decreases in proportion to the square of the distance ratio. It is essential to remember that this law is only applicable under certain conditions where the wave can spread out uniformly from the source.
Energy Rate Calculation
Energy rate is the measure of how much energy passes through a certain area in a given time. It is commonly referred to as power and it is calculated using the wave's intensity and the area the wave passes through. The formula is:
\[P = I \times A\]where:
  • \( P \) is the power in joules per second (J/s), which is also the energy rate
  • \( I \) is the intensity of the wave in joules per square meter per second (J/m虏路s)
  • \( A \) is the area through which the wave passes, in square meters (m虏)
This calculation helps us understand how much energy the wave is transferring through a specific area. It is especially useful in practical applications such as determining the impact force a wave might have on structures. By finding the energy rate, one can make informed decisions about safety and design requirements.
Wave Intensity Formula
Wave intensity is a crucial concept in understanding wave propagation, particularly in the context of earthquake waves. The intensity of the wave provides an insight into the amount of energy being transferred by the wave over a unit area per unit time. The formula for intensity is:
\[I = \frac{P}{A}\]where:
  • \( I \) is the wave intensity
  • \( P \) is the power or energy rate in J/s
  • \( A \) is the area through which the wave energy is passing, in m虏
Understanding wave intensity allows us to evaluate the strength of earthquake waves at different locations. By using this formula, scientists can estimate the potential impact and intensity of seismic waves, which is valuable for predicting and mitigating earthquake effects.

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Most popular questions from this chapter

(II) A satellite dish is about \(0.5 \mathrm{~m}\) in diameter. According to the user's manual, the dish has to be pointed in the direction of the satellite, but an error of about \(2^{\circ}\) to either side is allowed without loss of reception. Estimate the wavelength of the electromagnetic waves (speed \(=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) received by the dish.

(1) In an earthquake, it is noted that a footbridge oscillated up and down in a one-loop (fundamental standing wave) pattern once every 1.5 \(\mathrm{s}\) . What other possible resonant periods of motion are there for this bridge? What frequencies do they correspond to?

(II) A \(65-\mathrm{cm}\) guitar string is fixed at both ends. In the frequency range between 1.0 and \(2.0 \mathrm{kHz},\) the string is found to resonate only at frequencies \(1.2,1.5,\) and \(1.8 \mathrm{kHz} .\) What is the speed of traveling waves on this string?

(II) (a) Show that the average rate with which energy is transported along a cord by a mechanical wave of frequency \(f\) and amplitude \(A\) is $$\bar{P}=2 \pi^{2} \mu v f^{2} A^{2}$$ where \(v\) is the speed of the wave and \(\mu\) is the mass per unit length of the cord. (b) If the cord is under a tension \(F_{\mathrm{T}}=135 \mathrm{~N}\) and has mass per unit length \(0.10 \mathrm{~kg} / \mathrm{m},\) what power is required to transmit \(120-\mathrm{Hz}\) transverse waves of amplitude \(2.0 \mathrm{~cm} ?\)

(II) \(\mathrm{A} .0 .40\) -kg cord is stretched between two supports, 7.8 \(\mathrm{m}\) apart. When one support is struck by a hammer, a transverse wave travels down the cord and reaches the other support in 0.85 \(\mathrm{s}\) . What is the tension in the cord?
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