![]() ![]() The period of a sound wave is typically measured in milliseconds. For typical sound waves, the maximum displacement of the molecules in the air is only a hundred or a thousand times larger than the molecules themselves - and what technologies are there for tracking individual molecules anyway? The velocity and acceleration changes caused by a sound wave are equally hard to measure in the particles that make up the medium.ĭensity fluctuations are minuscule and short lived. Measuring displacement might as well be impossible. The density amplitude is the maximum change in density.The pressure amplitude is the maximum change in pressure (the maximum gauge pressure).Amplitudes associated with changes in bulk properties of arbitrarily small regions of the medium.The acceleration amplitude is the maximum change in acceleration.The velocity amplitude is the maximum change in velocity.The displacement amplitude is the maximum change in position.Amplitudes associated with changes in kinematic quantities of the particles that make up the medium.The amplitude of a sound wave can be quantified in several ways, all of which are a measure of the maximum change in a quantity that occurs when the wave is propagating through some region of a medium. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License. Use the information below to generate a citation. Then you must include on every digital page view the following attribution: If you are redistributing all or part of this book in a digital format, Then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a print format, Want to cite, share, or modify this book? This book uses the This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Since the arc subtends an angle ϕ ϕ at the center of the circle, To calculate the intensity at an arbitrary point P on the screen, we return to the phasor diagram of Figure 4.7. In solving that problem, you will find that they are less than, but very close to, ϕ = 3 π, 5 π, 7 π, … rad. ![]() The exact values of ϕ ϕ for the maxima are investigated in Exercise 4.120. As a result, E 1 E 1 and E 2 E 2 turn out to be slightly larger for arcs that have not quite curled through 3 π 3 π rad and 5 π 5 π rad, respectively. Since the total length of the arc of the phasor diagram is always N Δ E 0, N Δ E 0, the radius of the arc decreases as ϕ ϕ increases. These two maxima actually correspond to values of ϕ ϕ slightly less than 3 π 3 π rad and 5 π 5 π rad. The proof is left as an exercise for the student ( Exercise 4.119). In part (e), the phasors have rotated through ϕ = 5 π ϕ = 5 π rad, corresponding to 2.5 rotations around a circle of diameter E 2 E 2 and arc length N Δ E 0. The amplitude of the phasor for each Huygens wavelet is Δ E 0, Δ E 0, the amplitude of the resultant phasor is E, and the phase difference between the wavelets from the first and the last sources is The phasor diagram for the waves arriving at the point whose angular position is θ θ is shown in Figure 4.7. This distance is equivalent to a phase difference of ( 2 π a / λ N ) sin θ. If we consider that there are N Huygens sources across the slit shown in Figure 4.4, with each source separated by a distance a/N from its adjacent neighbors, the path difference between waves from adjacent sources reaching the arbitrary point P on the screen is ( a / N ) sin θ. To calculate the intensity of the diffraction pattern, we follow the phasor method used for calculations with ac circuits in Alternating-Current Circuits. Calculate the intensity relative to the central maximum of an arbitrary point on the screen.Calculate the intensity relative to the central maximum of the single-slit diffraction peaks.By the end of this section, you will be able to: ![]()
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