Two sitar strings A and B 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?
The frequency of string B is 318 Hz.
Explanation
This problem tests the understanding of beats and the relationship between tension and frequency in a string. The key insight comes from the context: 'Increase in the tension of a string increases its frequency.' Therefore, reducing tension decreases frequency. By analyzing whether the beat frequency increases or decreases after the tension change, we can determine which string has the higher frequency.
Solution Steps
Step 1: Given: Original frequency of A (ν_A) = 324 Hz, Initial beat frequency = 6 Hz. Therefore, |ν_A - ν_B| = 6 Hz. So ν_B could be either 330 Hz or 318 Hz.
Step 2: When tension in string A is reduced, its frequency decreases (as stated in the context: increase in tension increases frequency, so decrease in tension decreases frequency). So new frequency ν_A' < 324 Hz.
Step 3: The beat frequency reduces from 6 Hz to 3 Hz after reducing A's frequency. If ν_B were 330 Hz (greater than ν_A), reducing ν_A would increase the difference |ν_B - ν_A'|, increasing the beat frequency. But beat frequency decreased.
Step 4: Therefore, ν_B must be less than ν_A. So ν_B = 324 - 6 = 318 Hz.
Step 5: Verification: Initially |324 - 318| = 6 Hz. After reducing A's frequency, the difference decreases to 3 Hz, which matches the given condition.