A tuning fork produces 6 beats per second with both, 60 cm and 40 cm of the stretched wire of sonometer. The frequency of fork is

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Let the frequency of the tuning fork be $n$ .

The frequency of a stretched sonometer wire is given by $f = \frac{v}{2 L}$ , which implies $f \propto \frac{1}{L}$ .

Let $f_{1}$ and $f_{2}$ be the frequencies of the wire of lengths $L_{1} = 60 cm$ and $L_{2} = 40 cm$ respectively.

Since $L_{1} > L_{2}$ , we have $f_{1} < f_{2}$ .

The tuning fork produces $6$ beats per second with both lengths. Thus, $n$ must lie between $f_{1}$ and $f_{2}$ .

$n - f_{1} = 6 \Rightarrow f_{1} = n - 6$

$f_{2} - n = 6 \Rightarrow f_{2} = n + 6$

Since $f \propto \frac{1}{L}$ , we have $f_{1} L_{1} = f_{2} L_{2}$ .

Substituting the values:

$(n - 6) \times 60 = (n + 6) \times 40$

$3 (n - 6) = 2 (n + 6)$

$3 n - 18 = 2 n + 12$

$n = 30 Hz$

Answer: $30 Hz$

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