Fundamental Wavelength of Vibration of Steel Wire 1m Apart

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The fundamental wavelength of a steel wire stretched taut between supports one meter apart is determined by the requirement that the wave has zero displacement at the fixed endpoints. The longest wavelength, or fundamental wavelength, for this setup is 2 meters, as it must have nodes at both ends. While some argue that a wavelength of 1 meter would require slack, it is clarified that a node can exist at the center of the wire without slack, provided the wire is under tension. The discussion also acknowledges that the wire can support higher frequency harmonics with shorter wavelengths. Understanding these principles is essential for analyzing the vibrational characteristics of the wire.
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A steel wire is stretched taut between supports one meter apart. What is the fundamental wavelength of vibration of the wire?
 
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Begin by defining "fundamental."

- Warren
 
Since the wave is fixed at the two poles, the wave must have 0 value at those two points. The standard sine wave (wave length 2pi) is 0 at 0, pi, and 2pi. A wave of wave length l will be 0 at 0, l/2 and l. In particular, the longest wave (which I take is what you mean by "fundamental wavelength") that is 0 at 0 and 1 must have wavelength 2.
 
I get it, so the only way for it to have a wavelength of 1 m is if there was a node in the wire, and then it would have to have some slack, which it doesnt.

Thanks.
 
No, it is quite possible for the wire to have a node in the center without "slack" (it has to be under tension and able to stretch in order to form waves). You asked for the fundamentalwavelength which is the longest wavelength (lowest frequency). It is possible for the wire to have arbitrarily small wavelength waves ("harmonics").
 

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