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# The frequency of a tuning fork $P$ is less by $1\%$ than the frequency of a standard tuning fork $S$ while the frequency of another tuning fork $Q$ is more by $2\%$ 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)$ $396Hz,408Hz$

Last updated date: 20th Sep 2024
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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 $1\%$ in the fork s.
Whereas in another have $2\%$ 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,
$\Rightarrow \dfrac{{{\delta _P}}}{{{\delta _S}}} = \dfrac{{99}}{{100}}$
And
$\Rightarrow \dfrac{{{\delta _Q}}}{{{\delta _S}}} = \dfrac{{102}}{{100}}$
As we know beats are equal to
$\Rightarrow {\delta _Q} - {\delta _P}$
Since the total number of the beat is 9
Therefore,
$\Rightarrow 9 = \dfrac{{102}}{{100}}{\delta _S} - \dfrac{{99}}{{100}}{\delta _S}$
Calculating for the value of ${\delta _S}$, we get
$\Rightarrow \dfrac{{9 \times 100}}{3} = {\delta _S}$
Therefore,
$\Rightarrow {\delta _S} = 300Hz$
Now
$\Rightarrow {\delta _P} = \dfrac{{99}}{{100}}{\delta _S}$
Again we will calculate the value for this, we get
$\Rightarrow \dfrac{{99}}{{100}} \times 300$
Therefore.
$\Rightarrow {\delta _P} = 297Hz$
Similarly,
$\Rightarrow {\delta _Q} = \dfrac{{102}}{{100}}{\delta _S}$
Now putting the values which we had calculated earlier, we get
$\Rightarrow \dfrac{{102}}{{100}} \times 300$
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.