The frequency of a stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. If the tension in the wire is increased by $8 N,$ it is in resonance with the first overtone of the closed tube. The initial tension in the wire is .... $N$
Diffcult
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(a) According to problem
$\frac{1}{{2L}}\sqrt {\frac{T}{m}} = \frac{v}{{4L}}$…..$(i) $
and $\frac{1}{{2L}}\sqrt {\frac{{T + 8}}{m}} = \frac{{3v}}{{4L}}$ ..…$(ii)$
Dividing equation $(i)$ and $(ii),$ $\sqrt {\frac{T}{{T + 8}}} = \frac{1}{3} \Rightarrow T = 1N$
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