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

The correct option is B 144 Hz

Let frequency turning forces be $x,x+6x+6\left(2\right)+....$

According to given frequency of $I^{st}\times 2=$ frequency of last

$2x=x+6\left(23\right)$

$x=138$

$\therefore$ frequency of $2^{nd}$ will be $x+6=144$