A wire under tension 225 N produces 6 beats | vedclass

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(B) Let $n$ be the frequency of the tuning fork.Let $n_1$ and $n_2$ be the frequencies of the wire at tensions $T_1 = 225 \ N$ and $T_2 = 256 \ N$ res

Vedclass · Accepted Answer

$186$

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(B) Let $n$ be the frequency of the tuning fork.Let $n_1$ and $n_2$ be the frequencies of the wire at tensions $T_1 = 225 \ N$ and $T_2 = 256 \ N$ respectively.Since the frequency of a stretched wire is proportional to the square root of tension $(n \propto \sqrt{T})$,we have:$n_1 = n - 6$ (or $n + 6$)$n_2 = n + 6$ (or $n - 6$)Since $T_2 > T_1$,it follows that $n_2 > n_1$. Thus,$n_1 = n - 6$ and $n_2 = n + 6$.Taking the ratio:$\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{225}{256}} = \frac{15}{16}$$\frac{n - 6}{n + 6} = \frac{15}{16}$$16(n - 6) = 15(n + 6)$$16n - 96 = 15n + 90$$n = 186 \ Hz$

$A$ wire under tension $225 \ N$ produces $6$ beats per second when it is tuned with a tuning fork. When the tension changes to $256 \ N$,it is again tuned with the same tuning fork,and the number of beats remains unchanged. The frequency of the tuning fork will be: (in $Hz$)

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