A tuning fork gives 5 beats per second with 40 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, $n^{'} \propto \frac{1}{l}$ .

Let $n_{1}$ and $n_{2}$ be the frequencies of the wire for lengths $l_{1} = 40$ cm and $l_{2} = 39$ cm respectively.

Since $l_{1} > l_{2}$ , we have $n_{1} < n_{2}$ .

Given that the number of beats is 5 in both cases, the frequency of the tuning fork must lie between $n_{1}$ and $n_{2}$ . Thus, we can write:

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

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

According to the law of length for a sonometer wire, $n_{1} l_{1} = n_{2} l_{2}$ .

Substituting the values:

$(n - 5) \times 40 = (n + 5) \times 39$

$40 n - 200 = 39 n + 195$

$40 n - 39 n = 195 + 200$

$n = 395$ Hz

Answer: 395 Hz

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