Question

Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies are

• A. 80 and 40
• B. 100 and 50
• C. 44 and 22
• D. 72 and 36

Answer 1

The correct option is D 72 and 36


Using $n_{Last}=n_{First}+\left(N-1\right)x$
where $N$= Number of tuning fork in series
$x=$ beat frequency between two successive forks
$\Rightarrow 2n=n+(10-1)\times 4\Rightarrow n=36\mathrm{Hz}$
$\therefore n_{First}=36\mathrm{Hz}\mathrm{and}{n}_{Last}=72\mathrm{Hz}$