Question

Ten forks are arranged in increasing order of frequency in such a way that any two consecutive tuning forks produce $4$ beats per second. The highest frequency is twice that of the lowest. Possible highest and the lowest frequency (in Hz) are
$\left( A \right)80\,and\,40$
$\left( B \right)100\,and\,150$
$\left( C \right)44\,and\,22$
$\left( D \right)72\,and\,36$

Answer 1

Each fork frequency has four bet per second given in the problem. Using that information we can calculate the frequency of the tenth fork. Now putting this frequency value in the highest and lowest frequency relation we can find the value of the lowest frequency. As the highest frequency is twice the lowest, multiply the lowest frequency with two we will get our solution.

Complete answer:
As per the given problem ten forks are arranged in increasing order of frequency in such a way that any two consecutive tuning forks produce $4$ beats per seconds and the highest frequency is twice of the lowest.
We need to calculate the value of the highest frequency and the lowest frequency.
As we know each fork produces $4$ beats per second with the previous one which means wach fork has a frequency $4Hz$ more than the previous.
Hence we can conclude that the beat frequency between two successive forks is $4Hz$.
So the tenth fork will have a frequency of
$\left( {N - 1} \right) \times 4Hz \ldots \ldots \left( 1 \right)$
Where,
N = Number of tuning fork in series
We know $N = 10$ as there are ten forks (Given)
Now putting the equation $\left( 1 \right)$ we will get,
$\left( {10 - 1} \right) \times 4Hz$
$\Rightarrow 9 \times 4Hz = 36Hz$
We know,
$v_higher = v_lower + \left( {N - 1} \right) \times x \ldots \ldots \left( 2 \right)$
This above formula is the relationship between the highest and the lowest frequency forks.
Where,
The higher frequency is equal to $v_higher$.
The lowest frequency is equal to $v_lower$.
Number of tuning forks in series is equal to$N = 10$.
The beat frequency between two successive forks is $x = 4Hz$.
From the problem we know that,
$v_higher = 2v_lower \ldots \ldots \left( 3 \right)$
Now putting all the values in the equation $\left( 2 \right)$ we will get,
$2v_lower = v_lower + \left( {10 - 1} \right) \times 4Hz$
Rearranging the above equation we will get,
$2v_lower - v_lower = \left( {10 - 1} \right) \times 4Hz$
$v_lower = 9 \times 4Hz$
Hence the value of the lowest frequency is
$v_lower = 36Hz$
Now putting this value in equation $\left( 3 \right)$ we will get,
$v_higher = 2 \times \left( {36Hz} \right)$
$\Rightarrow v_higher = 72Hz$

Therefore the correct option is $\left( D \right)$.

Note:
In place of $\left( {N - 1} \right) \times x$ in equation $\left( 2 \right)$ we can also directly put the value of the tenth fork frequency. And also be careful while calculating the frequency of the tenth fork, never multiply the beat frequency between two successive forks directly with ten because after the first frequency only, there is a consecutive increase in beat frequency. Therefore it multiplies with $N - 1$ terms.