A tuning fork gives 5 beats per second with a 33 cm length of sonometer wire. If the length of the wire is shortened by 1 cm, the number of beats is still the same. The frequency of the fork is

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

The frequency of a sonometer wire is inversely proportional to its length, $f \propto \frac{1}{L}$ .

Let $f_{1}$ and $f_{2}$ be the frequencies of the wire for lengths $L_{1} = 33$ cm and $L_{2} = 32$ cm respectively.

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

The tuning fork gives $5$ beats per second with both lengths. Thus, $f_{1}$ must be less than $n$ and $f_{2}$ must be greater than $n$ .

$n - f_{1} = 5 \Rightarrow f_{1} = n - 5$

$f_{2} - n = 5 \Rightarrow f_{2} = n + 5$

Using the relation $f_{1} L_{1} = f_{2} L_{2}$ :

$(n - 5) \times 33 = (n + 5) \times 32$

$33 n - 165 = 32 n + 160$

$n = 325$ Hz

Answer: $325$ Hz

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