The frequency of tuning forks $A$ and $B$ are respectively $3\%$ more and $2\%$ less than the frequency of…
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 Hz) is
98
100
103
105
Solution
Let $f$ be the frequency of fork $C$, then
$f_A = f + \frac{3f}{100} = \frac{103f}{100}$ and $f_B = f - \frac{2f}{100} = \frac{98f}{100}$
But $f_A - f_B = 5 \Rightarrow \frac{103f}{100} - \frac{98f}{100} = 5 \Rightarrow 5f = 500$
$\Rightarrow f = 100\text{ Hz}$
$\therefore f_A = \frac{(103)(100)}{100} = 103\text{ Hz}$