Two sitar strings A and B playing the note ‘Dha’ are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if the frequency of A is 427 Hz?
422 Hz
Explanation
This question tests the understanding of beats and the relationship between tension and frequency of a vibrating string. The key insight is that increasing tension increases the frequency of a string. By analyzing whether the beat frequency increases or decreases when tension is changed, we can determine whether the original frequency of B was higher or lower than A. Since the beat frequency decreased when B's tension was increased, B's frequency must have been lower than A's initially.
Solution Steps
Given: Beat frequency = 5 Hz, Frequency of A (v_A) = 427 Hz, After increasing tension of B, beat frequency = 3 Hz
Key Principle: Increase in the tension of a string increases its frequency.
Analysis: If v_B > v_A initially, increasing v_B would increase the beat frequency. But beat frequency decreased from 5 Hz to 3 Hz. This shows that v_B < v_A.
Calculation: Since v_B < v_A, we have: v_A - v_B = 5 Hz. Substituting v_A = 427 Hz: 427 - v_B = 5 Hz. Therefore, v_B = 427 - 5 = 422 Hz.