Question

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/sec are produced, then the frequency of the tuning fork A(in Hz) is

• A. 98
• B. 100
• C. 103
• D. 105

Answer 1

The correct option is C 103

Beats between $A$ and $B=n_A-n_B=5$ beats/sec
Given $A$ is $3$% more than $C$ and $B$ is $2$% less than $C$.
$\therefore A-B=5$ beats/sec
$\therefore (A-B)\propto 100$
$\therefore$ If $C$ is $100Hz$
$\Rightarrow$ $A$ is $3+100=103Hz$
$\Rightarrow B$ is $100-2=98Hz$