1. Tardigrade
  2. Question
  3. Physics
  4. 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

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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

Waves

Solution:

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
$2 v=v+138$ or $v=138$
$\therefore \,\,\,\,$ Frequency of $2^{\text {nd }}$ fork $=v+6=138+6=144 \,Hz$