Question

A sonometer wire 110cm long produces a resonance with a tuning fork. When its length is decreased by 10 cm, 9 beats per second are heard. The frequency of the tuning fork is.
$A.$ 90 Hz
$B.$ 85 Hz
$C.$ 82 Hz
$D.$ 75 Hz

Answer 1

- Hint: Number of beats per second heard by the vibrating source is equal to difference in frequencies of the vibrating source, also frequency is inversely proportional to length of vibrating source. Use these two relations to get the frequency of the fork.

Complete step-by-step solution -

As we know that, number of beats per second of the vibrating sources = difference in frequencies of the source.
So, the frequency of the fork is ${n_1}$ when the sonometer wire length is ${l_1}$ and the frequency of tuning fork is ${n_2}$, when length is ${l_2}$.
By using above formula,
| ${n_1} - {n_2}$ | = 9
Also as we know frequency is inversely proportional to length of vibrating source, so
${n_1}$ $\propto \dfrac{1}{{{l_1}}}$
${n_1}$ = $\dfrac{k}{{{l_1}}}$
here k is any constant,
similarly for ${n_2}$ also,
${n_2}$ = $\dfrac{k}{{{l_2}}}$
Now by dividing these two equations,
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{{l_2}}}{{{l_1}}}$
As we know that ${l_1}$ = 110 cm and ${l_2}$ = 100 cm so,
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{100}}{{110}}$
After cancelling 0 it becomes,
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{10}}{{11}}$
${n_1} = \dfrac{{10}}{{11}}{n_2}$
Now put value of ${n_1}$ in first equation,
| $\dfrac{{10}}{{11}}{n_2} - {n_2}$ | =9
After taking lcm,
| $\dfrac{{10{n_2} - 11{n_2}}}{{11}}$ | = 9
After subtracting,
| $\dfrac{{ - 1{n_2}}}{{11}}$ | =9
Since any quantity inside a mod comes out as positive outside mod so,
$\dfrac{{{n_2}}}{{11}}$ = 9
${n_2}$ =99
Put value of ${n_2}$ in second equation ,
$n{{\kern 1pt} _1} = \dfrac{{10}}{{11}} \times 99$
After dividing 99 by 11,
${n_1}$ = 10$\times$ 9
${n_1}$ =90
So option A is the correct answer.

Note : The frequency of the beats refers to the degree to which the volume is felt oscillating from the top to the low volume. For example, if two complete cycles of high and low values ​​are heard every second, the frequency is 2 Hz. The frequency of the beat remains equal to the difference in frequency of the two notes that interfere with the production of the beat.