And so the fundamental frequency can then be ricin as the over the fundamental wavelength. So we can say the fundamental wavelength would be equaling two times the length. This means that the length of the string is equal to half two a half wavelength on the fundamental resident's frequency.
That is the end of the solution.If the string is fixed at both ends. Newton's divided by 6.0 times 10 to the second Newtons close parentheses and this is giving us 5.64 hurts or we can say 5.64 beats for a second. This would be multiplied by one minus the square root UH, 5.40 times 10 to the second. So this would be F sub one multiplied by one minus the square root of prime divided by F. And so when the two strings are sounded together, this means that the beat frequency the frequency of the beat equals F sub one, the fundamental frequency of the first string, minus the fundamental frequency of the second string F So one crime. Tension in the second string divided by the force tension in the first string. And so this would simply be equaling to the fundamental frequency of the first shrink multiplied by the square root of the force. Ah, Jeff over for length squared mu and then multiplied by f prime divided by f, this algebraic lee makes sense. If a second, identical call string with attention f crime, this is going to be less than F if that struck that the fundamental frequency of that vibration would be f the fundamental frequency prime would be the square root of F crime divided by four I'll square and mu, and this is gonna be equaling to the square root. And so this would be equal in the square root of the force over four l squared, mute, and so in. This would be equaling 21 over two times the length multiplied by the square root of the tension force, divided by the linear density, the linear mass density.
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