32 tuning forks are arranged in descending order of frequencies. If an

1.	32 tuning forks are arranged in descending order of frequencies. If any two consecutive tuning forks are sounded together, the number of beats heard is eight per second. The frequency of the first tuning fork is octave of the last fork. Calculate the frequency of the first, last and the 21st fork.
Answer» Data : n₁ = 2n₃₂ , (n₁ is octave of n₃₂) beat frequency = 8 Hz The set of tuning forks is arranged in descending order of their frequencies. ∴ n₂ = n₁ – 8 n₃ = n₂ – 8 = n₁ – 2 × 8 n₄ = n₃ – 8 = n₁ – 3 × 8 ∴ n₃₂ = n₁ – 31 × 8 = n₁ – 248 Since n₁ = 2n₃₁ , n₃₂ = 2n₃₂ – 248 ∴ The frequency of the last fork, n₃₂ = 248 Hz The frequency of the first fork, n₁ = 2n₃₂= 2 × 248 = 496 Hz ∴ The frequency of the 21st fork, n₂₁ = n₁ – 20 × 8 = 496 -160 = 336 Hz

32 tuning forks are arranged in descending order of frequencies. If any two consecutive tuning forks are sounded together, the number of beats heard is eight per second. The frequency of the first tuning fork is octave of the last fork. Calculate the frequency of the first, last and the 21st fork.

Answer»

Data : n₁ = 2n₃₂ , (n₁ is octave of n₃₂) beat frequency = 8 Hz

The set of tuning forks is arranged in descending order of their frequencies.

∴ n₂ = n₁ – 8

n₃ = n₂ – 8 = n₁ – 2 × 8

n₄ = n₃ – 8 = n₁ – 3 × 8

∴ n₃₂ = n₁ – 31 × 8 = n₁ – 248

Since n₁ = 2n₃₁ , n₃₂ = 2n₃₂ – 248

∴ The frequency of the last fork, n₃₂ = 248 Hz

The frequency of the first fork, n₁ = 2n₃₂= 2 × 248 = 496 Hz

∴ The frequency of the 21st fork, n₂₁ = n₁ – 20 × 8 = 496 -160 = 336 Hz