/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 A two-dimensional water wave spr... [FREE SOLUTION] | 91Ó°ÊÓ

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A two-dimensional water wave spreads in circular ripples. Show that the amplitude \(A\) at a distance \(r\) from the initial disturbance is proportional to \(1 / \sqrt{r}\). (Suggestion: Consider the energy carried by one outward- moving ripple.)

Short Answer

Expert verified
The amplitude (A) of a two-dimensional water wave spreading in circular ripples is directly proportional to the inverse of the square root of the distance from the initial disturbance, \(A \propto \frac{1}{\sqrt{r}}\), because the intensity of a wave decreases with the square of the distance from the source due to the energy of the wave being spread over a larger area.

Step by step solution

01

Conceptualizing the Wave Expansion

First, envision the wave as a series of concentric circles, each being a ripple produced from the central disturbance. As the ripples move outwards, the total energy that was initially at the point of disturbance is shared among larger and larger circles. The magnitude of this energy remains constant but it is spread over a larger area as the ripple expands.
02

Define the Area Covered by the Wave

Considering that the waves are expanding in a circular format, the surface area they cover can be defined using the formula for the area of a circle, \(A = \pi r^2\), where \(r\) is the radius or distance from the center of disturbance to the location of the wave.
03

Derive the Amplitude

We know that the intensity (I) of a wave is its power (P) divided by the area (A) it covers. With P being a constant (since the energy remains constant), it implies that I is inversely proportional to A, hence \(I \propto \frac{1}{A}\). Conversely, the square of the amplitude (A) of a wave is directly proportional to its intensity, i.e. \(I \propto A^2\). Combining these two proportionalities, we have \(A^2 \propto \frac{1}{A}\). Taking the square root of both sides, we find that \(A \propto \frac{1}{\sqrt{r}}\) as we were required to prove.

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Key Concepts

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

Wave Energy Conservation
Energy conservation is an important principle in the study of waves, especially when analyzing how they propagate through a medium. Imagine you drop a pebble into a still pond; the energy transferred to the water from this initial disturbance causes waves to ripple outward. This energy must be conserved, meaning that it doesn't magically increase or disappear as the wave spreads—it just gets distributed over an expanding area.

As suggested in the exercise, we consider the energy carried by one moving ripple. Since energy is conserved, the total amount of energy present in the initial disturbance is the same as the energy distributed across the wave fronts as they expand. Mathematically, if the power of the wave remains constant, the intensity of the wave will change in response to the increasing area it has to cover. Therefore, we understand that despite the energy being constant, the larger the area, the more diluted the energy becomes, which affects the amplitude of the wave.
Circular Wave Expansion
The concept of circular wave expansion can be visualized as the ripples on a pond after the aforementioned pebble is dropped. These waves travel outward in concentric circles, each representing a wave front. The area of each subsequent circle increases as the wave moves further from the origin.

Calculating the area of these circles with the formula \(A = \pi r^2\) helps us understand that as the radius \(r\) gets larger, the area increases by the square of the radius. It implies that, at a larger distance, the same amount of wave energy spreads over a much greater area, effectively reducing the energy per unit area available to exert influence, which in practical terms, can be experienced as a decrease in the wave's amplitude.
Intensity of a Wave
The intensity of a wave is a measure of the power per unit area that is transmitted by the wave. It's a concept that can be somewhat counterintuitive since it involves the amount of energy passing through a unit area, not just the total energy in the wave.

By knowing that intensity is power divided by area (\(I = P/A\)), and considering that wave energy (and therefore power) is conserved, any increase in area results in a decrease in intensity. Intuitively, as a wave front expands outward, each point on the wave front has less energy passing through it per unit of time because the same amount of energy is now spread over a larger area.

Thus, connecting the dots, since the square of the amplitude is directly proportional to intensity (\(I \propto A^2\)), a reduction in intensity with increased distance from the source means that the amplitude must decrease correspondingly. When the amplitude's dependence on radius is calculated for a circular wave, it confirms the relationship \(A \propto 1/\sqrt{r}\), providing a mathematical explanation for our observations of wave behavior.

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Most popular questions from this chapter

A string on a musical instrument is held under tension \(T\) and extends from the point \(x=0\) to the point \(x=L .\) The string is overwound with wire in such a way that its mass per unit length \(\mu(x)\) increases uniformly from \(\mu_{0}\) at \(x=0\) to \(\mu_{L}\) at \(x=L .\) (a) Find an expression for \(\mu(x)\) as a function of \(x\) over the range \(0 \leq x \leq L\). (b) Show that the time interval required for a transverse pulse to travel the length of the string is given by $$\Delta t=\frac{2 L\left(\mu_{L}+\mu_{0}+\sqrt{\mu_{L} \mu_{0}}\right)}{3 \sqrt{T}(\sqrt{\mu_{L}}+\sqrt{\mu_{0}})}$$

S and P waves, simultaneously radiated from the hypocenter of an earthquake, are received at a seismographic station \(17.3 \mathrm{s}\) apart. Assume the waves have traveled over the same path at speeds of \(4.50 \mathrm{km} / \mathrm{s}\) and \(7.80 \mathrm{km} / \mathrm{s} .\) Find the distance from the seismograph to the hypocenter of the quake.

(a) Show that the function \(y(x, t)=x^{2}+v^{2} t^{2}\) is a solution to the wave equation. (b) Show that the function in part (a) can be written as \(f(x+v t)+g(x-v t),\) and determine the functional forms for \(f\) and \(g\). (c) What If? Repeat parts (a) and (b) for the function \(y(x, t)=\sin (x) \cos (v t)\).

A string of length \(L\) consists of two sections. The left half has mass per unit length \(\mu=\mu_{0} / 2,\) while the right has a mass per unit length \(\mu^{\prime}=3 \mu=3 \mu_{0} / 2 .\) Tension in the string is \(T_{0} .\) Notice from the data given that this string has the same total mass as a uniform string of length \(L\) and mass per unit length \(\mu_{0}\). (a) Find the speeds \(v\) and \(v^{\prime}\) at which transverse pulses travel in the two sections. Express the speeds in terms of \(T_{0}\) and \(\mu_{0},\) and also as multiples of the speed \(v_{0}=\left(T_{0} / \mu_{0}\right)^{1 / 2} .\) (b) Find the time interval required for a pulse to travel from one end of the string to the other. Give your result as a multiple of \(\Delta t_{0}=L / v_{0}\).

Two points \(A\) and \(B\) on the surface of the Earth are at the same longitude and \(60.0^{\circ}\) apart in latitude. Suppose that an earthquake at point \(A\) creates a \(P\) wave that reaches point \(B\) by traveling straight through the body of the Earth at a constant speed of \(7.80 \mathrm{km} / \mathrm{s} .\) The earthquake also radiates a Rayleigh wave, which travels across the surface of the Earth in an analogous way to a surface wave on water, at \(4.50 \mathrm{km} / \mathrm{s}\). (a) Which of these two seismic waves arrives at \(B\) first? (b) What is the time difference between the arrivals of the two waves at \(B ?\) Take the radius of the Earth to be \(6370 \mathrm{km}\).

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