Solution:
Since reducing the tension in B decreases its frequency to 630 Hz (unison with A), B's original frequency must have been higher than A's. Therefore, \(f_B = 630 + 3 = 633\text{ Hz}\).
In a guitar, two strings A and B are slightly out of tune and produce 3 beats per second. When tension in B is slightly reduced, both the strings come in unison. If frequency of A is 630 Hz, then original frequency of B was
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