NEET 2020 Physics Question 147 Solution

The beat frequency produced by two interfering acoustic sound waves is the absolute difference between their individual frequencies: $$f_{\text{beat}} = |f_A - f_B| = 6\text{ Hz}$$ Given that string A has a known frequency $$f_A = 530\text{ Hz}$$, there are two possible mathematical values for the initial frequency of string B ($$f_B$$): - Case 1: $$f_B = f_A + 6 = 530 + 6 = 536\text{ Hz}$$ - Case 2: $$f_B = f_A - 6 = 530 - 6 = 524\text{ Hz}$$

The fundamental frequency of a transverse wave on a stretched string of length $$L$$ and linear mass density $$\mu$$ is given by: $$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \implies f \propto \sqrt{T}$$ When the tension in string B is slightly decreased ($$T_B \downarrow$$), its frequency must definitely decrease ($$f_B' < f_B$$).

If $$f_B = 536\text{ Hz}$$, a decrease in tension causes $$f_B$$ to decrease (for instance, to $$535\text{ Hz}$$). Then: $$f_{\text{beat}}' = |530 - 535| = 5\text{ Hz}$$ The beat frequency would DECREASE. This directly contradicts the observed problem statement that the beat frequency INCREASES to 7 Hz.

If $$f_B = 524\text{ Hz}$$, a decrease in tension causes $$f_B$$ to decrease further away from $$f_A$$ (to $$523\text{ Hz}$$). Then: $$f_{\text{beat}}' = |530 - 523| = 7\text{ Hz}$$ The beat frequency INCREASES to 7 Hz, which perfectly matches the given experimental observation.

Therefore, the original frequency of string B was unequivocally 524 Hz.