A set of 24 tuning forks are so arranged that each gives 6 beats/s 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

A set of 24 tuning forks are so arranged that each gives 6 beats/s 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) 132 Hz (C) 144 Hz (D) 276 Hz

Tardigrade Admin · Accepted Answer

The frequencies of tuning forks are the terms of an A P whose common difference is 6 . ∴ l=a+(n-1) d 2 a=a+(24-1) × 6 a=23 × 6 =138 ∴ Second frequency =138+6=144 H z

Q. A set of $24$ tuning forks are so arranged that each gives $6$ beats/s 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

138 Hz

132 Hz

144 Hz

276 Hz

The frequencies of tuning forks are the terms of an $A P$ whose common difference is $6$ .
$∴ l = a + (n - 1) d$
$2 a = a + (24 - 1) \times 6$
$a = 23 \times 6$
$= 138$
$∴$ Second frequency
$= 138 + 6 = 144 Hz$

Q. A set of 24 tuning forks are so arranged that each gives 6 beats/s 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