Question

Two identical strings, A and B, have nearly the samtension. When they both vibrate in their fundamental resonantmodes, ther is a beat of 3Hz. When string B is tightenedslightly, to increase the tension, the beat frequency becomes 6 Hz.This means.

A)that before and after tightening B has a higher frequencythan that of AB)That before and after tightening A has a higher frequencythan B.C)that before tightening B had a higher frequency than A, butafter tightening, A has a higher frequency than B.d) that before tightening A had a higher frequecy than B, butafter tightening. B has a higher frequency than A.E) none of the above.

Answer 1

A. Before and after tightening B has a higher frequency than that of A

Step-by-step 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:

A. Before and after tightening B has a higher frequency than that of A