Two tuning forks when sounded together produce 4 beats per second. One of the forks is in unison with 23 cm length of sonometer wire and other with 24 cm length of the same wire. The frequencies of th

Question

Two tuning forks when sounded together produce 4 beats per second. One of the forks is in unison with 23 cm length of sonometer wire and other with 24 cm length of the same wire. The frequencies of the two tuning forks are

TestHub · Accepted Answer

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

According to the law of length for a sonometer wire, the frequency is inversely proportional to the resonating length, i.e., $n \propto \frac{1}{l}$ .

Therefore, $n_{1} l_{1} = n_{2} l_{2}$

Given $l_{1} = 23 cm$ and $l_{2} = 24 cm$ . Since $l_{1} < l_{2}$ , we have $n_{1} > n_{2}$ .

Substituting the values, we get:

$n_{1} \times 23 = n_{2} \times 24$

$n_{1} = \frac{24}{23} n_{2}$

The tuning forks produce 4 beats per second, so:

$n_{1} - n_{2} = 4$

Substituting the value of $n_{1}$ in the beat frequency equation:

$\frac{24}{23} n_{2} - n_{2} = 4$

$n_{2} (\frac{24}{23} - 1) = 4$

$n_{2} (\frac{1}{23}) = 4$

$n_{2} = 4 \times 23 = 92 Hz$

Now, finding $n_{1}$ :

$n_{1} = n_{2} + 4 = 92 + 4 = 96 Hz$

The frequencies of the two tuning forks are $96 Hz$ and $92 Hz$ .

Answer: $96 Hz, 92 Hz$

Physics - Wave on String Question with Solution | TestHub

Options:

Doubts & Discussion