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(D) The wave speed $v$ is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.01}} = 100\,m/s$.For a string fixed at both ends,the frequencies are

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$n_2 = \frac{f_1 + f_2}{2} = 75\,Hz$

Vedclass · Answer

(D) The wave speed $v$ is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.01}} = 100\,m/s$.For a string fixed at both ends,the frequencies are $f_n = \frac{nv}{2L} = \frac{n 	imes 100}{2 	imes 1} = 50n\,Hz$. Thus,$f_1 = 50\,Hz, f_2 = 100\,Hz, f_3 = 150\,Hz$.For a string fixed at one end,the frequencies are $n_k = \frac{(2k-1)v}{4L} = \frac{(2k-1) 	imes 100}{4 	imes 1} = 25(2k-1)\,Hz$ for $k=1, 2, 3, \dots$. Thus,$n_1 = 25\,Hz, n_2 = 75\,Hz, n_3 = 125\,Hz$.Comparing these,$n_2 = \frac{f_1 + f_2}{2} = \frac{50 + 100}{2} = 75\,Hz$.

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