A string is under tension of 180 N and mass | vedclass

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(D) The difference between two consecutive resonant frequencies of a stretched string is equal to the fundamental frequency $(f_0)$.$f_0 = f_2 - f_1 =

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$4$

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(D) The difference between two consecutive resonant frequencies of a stretched string is equal to the fundamental frequency $(f_0)$.$f_0 = f_2 - f_1 = 450 \ Hz - 375 \ Hz = 75 \ Hz$.The formula for the fundamental frequency is $f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T = 180 \ N$ and $\mu = 2 	imes 10^{-3} \ kg/m$.Substituting the values: $75 = \frac{1}{2L} \sqrt{\frac{180}{2 	imes 10^{-3}}} = \frac{1}{2L} \sqrt{90000} = \frac{300}{2L} = \frac{150}{L}$.Thus,$L = \frac{150}{75} = 2 \ m$.The total mass of the string is $m = \mu 	imes L = (2 	imes 10^{-3} \ kg/m) 	imes (2 \ m) = 4 	imes 10^{-3} \ kg = 4 \ g$.

$A$ string is under tension of $180 \ N$ and mass per unit length $2 \times 10^{-3} \ kg/m$. It produces two consecutive resonant frequencies with a tuning fork,which are $375 \ Hz$ and $450 \ Hz$. The mass of the string is: (in $g$)

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