Answer:
A. Before and after tightening B has a higher frequency than that of A
Explanation:
Since, the beat frequency is 3 Hz before tightening. Therefore, the difference between the frequencies of strings, A and B is 3 Hz. As, we tighten the string B, its tension increases. From the following relation it is clear that the increase in tension increases the fundamental frequency.
f₁ = (1/2L)√[T/μ]
where,
f₁ = fundamental frequency
L = length of string
T = Tension in string
μ = linear mass density of string
Therefore, frequency of B increases slightly. And the beat frequency or the difference between frequencies of A and B, becomes 6 Hz.
Now, there can be two cases, which are:
Either, the frequency of B was 3 Hz less than frequency of A earlier and after tightening frequency of B became 6 Hz greater than frequency of A.
Or, the frequency of B was 3 Hz greater than the frequency of A, and after tightening it became 6 Hz greater than frequency of A.
The second option seems more reasonable because it requires 3 Hz of increase in frequency of B, while the first option requires increase of 9 Hz in frequency of B. Since, the there is slight tightening. Therefore, the frequency will also increase by a slight value. Thus, the correct option is:
<u>A. Before and after tightening B has a higher frequency than that of A</u>
