When two tuning forks are sounded together, 6 beats per second are heard. One of the fork is in unison with 0.70 m length of sonometer wire and another fork is in unison with 0.69 m length of the same

Question

When two tuning forks are sounded together, 6 beats per second are heard. One of the fork is in unison with 0.70 m length of sonometer wire and another fork is in unison with 0.69 m length of the same sonometer wire. The frequencies of the two tuning forks are

TestHub · Accepted Answer

Let the frequencies of the two tuning forks be $n_{1}$ and $n_{2}$ .

The frequency of a sonometer wire is inversely proportional to its resonating length, so $n \propto \frac{1}{l}$ .

Given $l_{1} = 0.70 m$ and $l_{2} = 0.69 m$ .

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

The beat frequency is given as $n_{2} - n_{1} = 6 \Rightarrow n_{2} = n_{1} + 6$ .

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

Substituting the values:

$n_{1} \times 0.70 = (n_{1} + 6) \times 0.69$

$70 n_{1} = 69 n_{1} + 414$

$n_{1} = 414 Hz$

$n_{2} = 414 + 6 = 420 Hz$

The frequencies of the two tuning forks are $414 Hz$ and $420 Hz$ .

Answer: $414 Hz, 420 Hz$

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