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(D) The resonance frequencies of a string fixed at both ends are given by $f_n = n f_1$,where $f_1$ is the fundamental frequency.The difference betwee

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$8 	imes 10^{-3} \, kg$

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(D) The resonance frequencies of a string fixed at both ends are given by $f_n = n f_1$,where $f_1$ is the fundamental frequency.The difference between two consecutive resonance frequencies is equal to the fundamental frequency:$f_{n+1} - f_n = f_1 = 450 \, Hz - 375 \, Hz = 75 \, Hz$.The fundamental frequency is $f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length of the string,$T$ is the tension,and $\mu$ is the linear mass density.Substituting the values:$75 = \frac{1}{2L} \sqrt{\frac{360}{4 	imes 10^{-3}}}$$75 = \frac{1}{2L} \sqrt{90000} = \frac{300}{2L}$$L = \frac{300}{150} = 2 \, m$.The total mass of the string $m$ is given by $m = \mu 	imes L$.$m = (4 	imes 10^{-3} \, kg/m) 	imes (2 \, m) = 8 	imes 10^{-3} \, kg$.

$A$ string with a mass density of $4 \times 10^{-3} \, kg/m$ is under a tension of $360 \, N$ and is fixed at both ends. One of its resonance frequencies is $375 \, Hz$. The next higher resonance frequency is $450 \, Hz$. The mass of the string is:

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