The frequency of tuning forks $A$ and $B$ are respectively 3% more and 2% less than the frequency of tuning fork $C.$ When $A$ and $B$ are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork $'A'$ in (in $Hz$) is

Question

Dr. Abubacker · Accepted Answer

(c) Let $n$ be the frequency of fork $C$ then
$n_{A} = n + \frac{3n}{100} = \frac{103n}{100}$ and $n_{B} = n - \frac{2n}{100} = \frac{98n}{100}$
But $n_{A} - n_{B} = 5 \Rightarrow \frac{5n}{100} = 5 \Rightarrow n = 100\ Hz$
$	herefore n_{A} = \frac{(103)(100)}{100} = 103\ Hz$

Q. The frequency of tuning forks $A$ and $B$ are respectively 3% more and 2% less than the frequency of tuning fork $C.$ When $A$ and $B$ are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork $'A'$ in (in $Hz$) is

Explanation

Reviewed by Dr. Abubacker

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