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A tuning fork arrangement (pair) produces 4 beats / sec with one fork of frquency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats /sec. Then what is the frequency of the unknown fork.?
The fact that placing wax decreases the frequency of the unknown fork and also the beatfrequency states that the unknown fork is of higher frequency.
n - 288 = 4 ⇒ n = 292 cps
Remember whenever the beats decrease while you wax an unknown tuning fork then it is sure that the frequency of the tunig fork is more than that of the other . So just add the beats which you get before waxing to the other tuning fork''s frequency. ie., 288+4=292.
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