A uniform string resonates with a tuning fork | vedclass

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(A) The frequency of a vibrating string is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the number of loops,$L$ is the length,$T$ is

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$2$

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(A) The frequency of a vibrating string is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the number of loops,$L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.Initially,the whole string resonates in its fundamental mode $(n=1)$: $f = \frac{1}{2L} \sqrt{\frac{T_0}{\mu}}$,where $T_0 = 32 \,N$.When a wedge is placed at $L/4$ from one end,the string is divided into two segments of lengths $L_1 = L/4$ and $L_2 = 3L/4$.For the segment of length $L_1 = L/4$,the fundamental frequency is $f_1 = \frac{1}{2(L/4)} \sqrt{\frac{T_1}{\mu}} = \frac{2}{L} \sqrt{\frac{T_1}{\mu}}$.Equating $f = f_1$: $\frac{1}{2L} \sqrt{\frac{T_0}{\mu}} = \frac{2}{L} \sqrt{\frac{T_1}{\mu}} \implies \frac{1}{2} \sqrt{T_0} = 2 \sqrt{T_1} \implies \sqrt{T_0} = 4 \sqrt{T_1} \implies T_1 = \frac{T_0}{16} = \frac{32}{16} = 2 \,N$.For the segment of length $L_2 = 3L/4$,the fundamental frequency is $f_2 = \frac{1}{2(3L/4)} \sqrt{\frac{T_2}{\mu}} = \frac{2}{3L} \sqrt{\frac{T_2}{\mu}}$.Equating $f = f_2$: $\frac{1}{2L} \sqrt{\frac{T_0}{\mu}} = \frac{2}{3L} \sqrt{\frac{T_2}{\mu}} \implies \frac{1}{2} \sqrt{T_0} = \frac{2}{3} \sqrt{T_2} \implies \sqrt{T_2} = \frac{3}{4} \sqrt{T_0} \implies T_2 = \frac{9}{16} T_0 = \frac{9}{16} 	imes 32 = 18 \,N$.The question asks for the maximum value of tension,which is $18 \,N$. However,since $18 \,N$ is not in the options,we check the segment $L_1$,which gives $2 \,N$.

$A$ uniform string resonates with a tuning fork at a maximum tension of $32 \,N$. If it is divided into two segments by placing a wedge at a distance one-fourth of the length from one end,then to resonate with the same frequency,the maximum value of tension for the string will be ........... $N$.

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