Sound WavesEasy
Question
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. The frequency of the unknown fork is
Options
A.286 cps
B.292 cps
C.294 cps
D.288 cps
Solution
The fact that placing wax decreases the frequency of the unknown fork and also the beat frequency states that the unknown fork is of higher frequency
n - 288 = 4 ⇒ n = 292 cps
n - 288 = 4 ⇒ n = 292 cps
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