Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
11
Vibration and Waves
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11-1 to 11-3: Simple Harmonic Motion
11-4: Simple Pendulum
11-7 and 11-8: Waves
11-9: Energy Transported by Waves
11-11: Interference
11-12: Standing Waves; Resonance
11-13: Refraction
11-14: Diffraction

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 52
Q

The speed of waves on a string is 97 m/s. If the frequency of standing waves is 475 Hz, how far apart are two adjacent nodes?

A
0.10 m0.10 \textrm{ m}
Giancoli 7th Edition, Chapter 11, Problem 52 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Adjacent nodes and a waveform are half a wavelength apart. So, from this node to this node is one full wavelength because the wave is, you know, identical of these two points in the sense that it's going, the slope is upwards to the right and upwards to the right as well as having zero displacement. Whereas this point slope is down to the right and so it's not the same as this point, it's not congruent. So, the wavelength goes from one node not to the next node but to the one after that. So, the distance between adjacent nodes then is half the wavelength. And we have this formula that says that the speed of a wave is this frequency times this wavelength. And we can divide both sides by f. And that gives wavelength is v over f and then knowing that this distance, l, we're looking for between, distance between adjacent nodes is wavelength over 2 we have 1/2 times v over f. So it's 1/2 times 97 meters per second divided by 475 hertz which is 0.10 meters.

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