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Chapter 15: Problem 15

Why do circular water waves on the surface of a pond decrease in amplitude as they travel away from the source?

Short Answer

Expert verified
Answer: The amplitude of circular water waves decreases as they travel away from the source because the energy of the waves is distributed over expanding circular wavefronts. As the wavefronts expand in radius, they cover a larger surface area, causing the energy density to decrease, which in turn leads to a decrease in amplitude.

Step by step solution

01

Define amplitude

Amplitude refers to the maximum displacement of a wave from its equilibrium position. It is a measure of the wave's energy, as a wave with a higher amplitude has more energy than a wave with a lower amplitude.
02

Understand the circular wavefront

A circular wavefront is produced when a disturbance (e.g., a stone thrown into a pond) occurs at a single point on the surface of the water. The energy from this disturbance produces circular wavefronts, centered on the point of disturbance. As the wavefronts travel away from the source, they expand in a concentric and roughly circular pattern.
03

Explain energy distribution in circular wavefront

The energy from the disturbance is spread evenly over the surface of each successive wavefront. As a wavefront expands, its surface area increases; specifically, the surface area of a wavefront is proportional to the square of its radius (A = 4πr^2). Consequently, at a greater distance from the source, the same amount of energy is distributed over a larger area, resulting in a reduced energy density.
04

Relate energy density to amplitude

Since the amplitude of a wave is directly related to its energy, a decrease in energy density leads to a decrease in amplitude. As the wavefronts expand and travel away from the source, they encounter a larger area to cover. This causes the energy to be spread thinner, leading to a reduction in the amplitude of the waves.
05

Conclusion

Circular water waves on the surface of a pond decrease in amplitude as they travel away from the source because the energy of the waves is distributed over expanding circular wavefronts. As the wavefronts expand in radius, they cover a larger surface area, causing the energy density to decrease, which in turn leads to a decrease in amplitude.

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Most popular questions from this chapter

The displacement from equilibrium caused by a wave on a string is given by \(y(x, t)=(-0.00200 \mathrm{~m}) \sin \left[\left(40.0 \mathrm{~m}^{-1}\right) x-\right.\) \(\left.\left(800 . \mathrm{s}^{-1}\right) t\right] .\) For this wave, what are the (a) amplitude, (b) number of waves in \(1.00 \mathrm{~m},\) (c) number of complete cycles in \(1.00 \mathrm{~s},\) (d) wavelength, and (e) speed?

Two steel wires are stretched under the same tension. The first wire has a diameter of \(0.500 \mathrm{~mm}\), and the second wire has a diameter of \(1.00 \mathrm{~mm}\). If the speed of waves traveling along the first wire is \(50.0 \mathrm{~m} / \mathrm{s}\), what is the speed of waves traveling along the second wire?

The equation for a standing wave on a string with mass density \(\mu\) is \(y(x, t)=2 A \cos (\omega t) \sin (\kappa x) .\) Show that the average kinetic energy and potential energy over time for this wave per unit length are given by \(K_{\text {ave }}(x)=\mu \omega^{2} A^{2} \sin ^{2} \kappa x\) and \(U_{\text {ave }}(x)=T(\kappa A)^{2}\left(\cos ^{2} \kappa x\right)\)

Suppose that the tension is doubled for a string on which a standing wave is propagated. How will the velocity of the standing wave change? a) It will double. c) It will be multiplied by \(\sqrt{2}\). b) It will quadruple. d) It will be multiplied by \(\frac{1}{2}\).

Students in a lab produce standing waves on stretched strings connected to vibration generators. One such wave is described by the wave function \(y(x, t)=(2.00 \mathrm{~cm}) \sin \left[\left(20.0 \mathrm{~m}^{-1}\right) x\right] \cos \left[\left(150 . \mathrm{s}^{-1}\right) t\right],\) where \(y\) is the transverse displacement of the string, \(x\) is the position along the string, and \(t\) is time. Rewrite this wave function in the form for a right- moving and a left-moving wave: \(y(x, t)=\) \(f(x-v t)+g(x+v t)\); that is, find the functions \(f\) and \(g\) and the speed, \(v\)
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