Two sitar strings, A and B, are playing the note 'Ga,' are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced, and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

Explanation

The beat frequency is given by the difference in frequency between the two strings:f_beat = |fA - fB|We are given that the original beat frequency is 6 Hz and that when the tension in string B is reduced, the beat frequency is reduced by 3 Hz. This means that the new beat frequency is 6 - 3 = 3 Hz.Using the above equation, we can set up the following system of equations:6 = |324 - fB|3 = |324 - (fB - x)|where x is the change in frequency caused by the reduction in tension in string B.Solving the first equation for |324 - fB|, we get:|324 - fB| = 6which gives us two possible values for f_B:fB = 330 Hz or fB = 318 HzNext, we can use the second equation to determine which value of f_B is correct. Substituting in the first possible value, we get:|324 - 330| = 6 = |324 - (330 - x)|Simplifying, we get:x = 12This means that the original frequency of string B was:fB = 324 - 6 = 318 HzTherefore, the correct answer is D) 318.