Question

A set of $24$ tuning forks are so arranged that each gives $6$ beats per second with the previous one. If the frequency of the last tuning fork is double that of the first, frequency of the second tuning fork is

• A. 138 Hz
• B. 144 Hz
• C. 132 Hz
• D. 276 Hz

Answer 1

Answer: (B)

Let the frequency of $1^{\text{st }}$ fork be $v$. Then frequency of $2^{\text{nd }}$ fork $=v+6=v+6(2-1)$ and so frequency of $24^{\text{th }}$ fork $=v+6(24-1)=v+138$
As per given condition
$2v=v+138$ or $v=138$
$\therefore$ Frequency of $2^{\text{nd }}$ fork $=v+6=138+6=144Hz$