Question

Two strings X and Y of a sitar produce a beat of frequency 4Hz. When the tension of string y is slightly increased the beat frequency is found to be 2 Hz. If the frequency of X is 300 Hz, then the original frequency of y was?
$\begin{aligned} & A.296Hz \\\ & B.298Hz \\\ & C.302Hz \\\ & D.304Hz \\\ \end{aligned}$

Answer 1

We know that as the tension in string y increases, then its frequency should also increase. When the beat frequency decreases from 4Hz to 2Hz because the frequency is directly proportional to the square root of tension in the string. Use this understanding and its mathematical equation to solve for the required answer.
Formula used: $f\propto \sqrt{T}$

Complete answer:
We are provided in the question that there are two strings X and Y producing a beat frequency of 4Hz. And if the tension in the string y is increased slightly then its beat frequency also increases as 2Hz. This can be represented mathematically as:
${{f}_{x}}-{{f}_{y}}=4Hz.....(1)$
Where ${{f}_{x}}$ is representing the frequency of string x and ${{f}_{y}}$ is representing the frequency of string y.
$f\propto \sqrt{T}$
Where f is representing the frequency and T is representing the tension in the string. According to the question, when beat frequency decreases from 4Hz to 2Hz then the tension in the string y increases slightly by using the above relation between f and t. frequency of string y starts increasing. We know, frequency of string x that is ${{f}_{x}}=300Hz$ and we have to find the frequency of string y that is ${{f}_{y}}$
By putting ${{f}_{x}}$ in equation (1) we get,
$\begin{aligned} & 300-{{f}_{y}}=4 \\\ & {{f}_{y}}=300-4 \\\ & {{f}_{y}}=296Hz \\\ \end{aligned}$

Hence the correct option is (a) .

Note:
It is pointed out in this question that frequency and tension are proportional to each other. When frequency increases then tension also starts increasing and vice versa. Hertz is the unit of frequency and one hertz is equal to one cycle per second.