A string fixed at both ends resonates at a ce | vedclass

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(A) The fundamental frequency of a string fixed at both ends is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tens

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Tension is made four times and length is doubled

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(A) The fundamental frequency of a string fixed at both ends is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.For option $A$: $n' = \frac{1}{2(2l)} \sqrt{\frac{4T}{m}} = \frac{1}{4l} \cdot 2 \sqrt{\frac{T}{m}} = n$ (No change).For option $B$: $n' = \frac{1}{2(l/2)} \sqrt{\frac{2T}{m}} = \sqrt{2} n$ (Changes).For option $C$: $n' = \frac{1}{2(l/2)} \sqrt{\frac{T/2}{m}} = \frac{1}{\sqrt{2}} n$ (Changes).For option $D$: $n' = \frac{1}{2(2l)} \sqrt{\frac{2T}{m}} = \frac{1}{\sqrt{2}} n$ (Changes).Thus,the adjustment in option $A$ does not affect the fundamental frequency.

$A$ string fixed at both ends resonates at a certain fundamental frequency. Which of the following adjustments would not affect the fundamental frequency?

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