Two sitar strings \(\mathrm{A}\) and \(\mathrm{B}\) playing the note "Dha" are slightly out of time and produce beats of frequency \(5 \mathrm{~Hz}\). The tension of the string B is slightly increased and the beat frequency is found to decrease to \(3 \mathrm{~Hz}\). What is the original frequency of \(\mathrm{B}\) if the frequency of \(\mathrm{A}\) is \(427 \mathrm{~Hz}\) ? (A) 432 (B) 422 (C) 437 (D) 417

From the given problem, we have the following information: - Beat Frequency_1 = 5 Hz (before increasing the tension of string B) - Beat Frequency_2 = 3 Hz (after increasing the tension of string B) - Frequency of String A (f_A) = 427 Hz - Frequency of String B (f_B) =?

Before we proceed, it's essential to know how the beat frequency is related to the two frequencies producing the beats. So, we have: Beat Frequency = |Frequency of String A - Frequency of String B| Now we can write the equation for both given cases. - For the initial case, beat frequency is 5 Hz, so: Beat Frequency_1 = |f_A - f_B| - For the second case, beat frequency is 3 Hz after increasing the tension of string B, so the frequency of string B increases (f_B1 = new frequency of B): Beat Frequency_2 = |f_A - f_B1|

Now we substitute the given data into the equations we formulated and solve: 1) Equation for initial case: 5 Hz = |427 Hz - f_B| 2) Equation for the case after increasing the tension of string B: 3 Hz = |427 Hz - f_B1| Since the tension of string B is increased the frequency increases, we have f_B1 = f_B + x, where x is the frequency increase.

Let's rewrite equation 2 incorporating f_B1 = f_B + x. 3 Hz = |427 Hz - (f_B + x)| We have two unknowns (f_B and x) and two equations, we can solve for f_B. 1) 5 Hz = 427 Hz - f_B OR 5 Hz = f_B - 427 Hz Either f_B = 427 Hz + 5 Hz = 432 Hz (Case 1) or f_B = 427 Hz - 5 Hz = 422 Hz (Case 2) Now let's check, which case holds for the equation with the tension increase: Remember, x is a positive number since the frequency increased when we increased the tension. 2) 3 Hz = 427 Hz - (f_B + x) OR 3 Hz = (f_B + x) - 427 Hz For Case 1: If f_B = 432 Hz, we get x = -2 Hz, which is wrong (x should be positive). For Case 2: If f_B = 422 Hz, we get x = 2 Hz, which is correct (x is positive). So, the original frequency of string B is 422 Hz. Answer: (B) 422