A string under a tension of $129.6\,\, N$ produces $10\,\, beats /sec$ when it is vibrated along with a tuning fork. When the tension is the string is increased to $160\,\, N,$ it sounds in unison with same tuning fork. calculate fundamental freq. of tuning fork .... $Hz$
Diffcult
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frequency of tuning fork $ = \frac{{\sqrt {\frac{{160}}{{\rm{m}}}} }}{{2l}}$
According to the question
$\frac{1}{{2l}}\sqrt {\frac{{160}}{{\rm{m}}}} - \frac{1}{{2l}}\sqrt {\frac{{129.6}}{{\rm{m}}}} = 10$
$\frac{1}{{2l}}\sqrt {\frac{{10}}{m}} [4 - 3.6] = 10 \Rightarrow \frac{1}{{2l}}\sqrt {\frac{{10}}{m}} = 25$
so ${{\rm{V}}_{{\rm{TF}}}} = \frac{1}{{2l}}\sqrt {\frac{{160}}{{\rm{m}}}} = 4\left[ {\frac{1}{{2l}}\sqrt {\frac{{10}}{{\rm{m}}}} } \right] = 100\,{\rm{Hz}}$
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