Question

The frequency of a tuning fork $P$ is less by $1\mathrm{\%}$ than the frequency of a standard tuning fork $S$ while the frequency of another tuning fork $Q$ is more by $2\mathrm{\%}$ the frequency of$S$. If $9\text{ beats/s}$ are produced when $P$ and $Q$ are sounded together, the frequency of $P$ and $Q$ are respectively:
$\left(a\right)$ $148.5Hz,153Hz$
$\left(b\right)$ $198Hz,204Hz$
$\left(c\right)$ $297Hz,306Hz$
$\left(d\right)$ $396Hz,408Hz$

Answer 1

Hint: Tuning forks permit us to review the fundamental qualities of sound first hand. An implement emits a pure musical tone (after waiting a moment) because it vibrates when you strike it. There are two basic qualities of sound: One is the Pitch which will be high and low and the other one is the volume which may be loud and soft.
Formula used:
Number of beats,
$\Rightarrow {\delta }_Q-{\delta }_P$
Where, ${\delta }_Q\text{ and }{\delta }_P$are the beats of $Q$and $P$respectively.

Complete step by step solution:
Here in the question since there is a decrease in tuning fork frequency by $1\mathrm{\%}$ in the fork s.
Whereas in another have $2\mathrm{\%}$ more frequency. So we can write it as $99$ and $102$ as we had taken this after removing the percentage. For details see the below it will make more clear.
So from the above, we can make a mathematical equation which will be like this,
According to the question,
$\Rightarrow {\displaystyle \frac{{\delta }_P}{{\delta }_S}}={\displaystyle \frac{99}{100}}$
And
$\Rightarrow {\displaystyle \frac{{\delta }_Q}{{\delta }_S}}={\displaystyle \frac{102}{100}}$
As we know beats are equal to
$\Rightarrow {\delta }_Q-{\delta }_P$
Since the total number of the beat is 9
Therefore,
$\Rightarrow 9={\displaystyle \frac{102}{100}}{\delta }_S-{\displaystyle \frac{99}{100}}{\delta }_S$
Calculating for the value of ${\delta }_S$, we get
$\Rightarrow {\displaystyle \frac{9\times 100}{3}}={\delta }_S$
Therefore,
$\Rightarrow {\delta }_S=300Hz$
Now
$\Rightarrow {\delta }_P={\displaystyle \frac{99}{100}}{\delta }_S$
Again we will calculate the value for this, we get
$$\Rightarrow {\displaystyle \frac{99}{100}}\times 300$$
Therefore.
$\Rightarrow {\delta }_P=297Hz$
Similarly,
$\Rightarrow {\delta }_Q={\displaystyle \frac{102}{100}}{\delta }_S$
Now putting the values which we had calculated earlier, we get
$$\Rightarrow {\displaystyle \frac{102}{100}}\times 300$$
Therefore,
$\Rightarrow {\delta }_Q=306Hz$

Hence, the option $\left(c\right)$ is the required frequency, which is $297Hz,306Hz$.

Note: A tuning fork could be a sound resonator that could be a two-pronged fork. The prongs, known as tines, are made of a U-shaped bar of metal (usually steel). This bar of metal will move freely. It resonates at a selected constant pitch once set moving by putting it against an object. It sounds a pure musical tone when waiting for a flash to permit some high overtone sounds to die out.