A set of 12 tuning forks is arranged in order of increasing frequencie

1.	A set of 12 tuning forks is arranged in order of increasing frequencies. Each fork produces V beats per second with the previous one. The last is an octave of the first. The fifth fork has a frequency of 90 Hz. Find V and the frequency of the first and the last tuning forks.
Answer» Data : n_{i + 1} – n_i = Y, n₁₂ = 2n₁ , n₅= 90 Hz n₂ – n₁ = Y beats/s ∴ n₂ = n₁ + Y beats/s Similarly, n₃ = n₂ + Y = n₁+ Y + Y ∴ n₃ = n₁ + 2Y = n₁ + (3 – 1) Y ∴ n_x = n₁ + (x – 1) Y Similarly, n₁₂ = n₁ + (12 – 1) Y = n₁ + 11Y ∴ n₁₂ = 2n₁ = n₁ + 11Y ∴ n₁ = 11Y Also, n₅ = n₁ + (5 – 1) Y = n₁ + 4Y ∴ n₅ = 11Y + 4Y = 15Y ∵ n₅ = 90 Hz ∴ 15Y = 90 ∴ Y = 6 ∴ n₁ = 11Y beats/s = 11 × 6 beats/s = 66 Hz and n₁₂ = 2n₁ = 2 (66) = 132 Hz

A set of 12 tuning forks is arranged in order of increasing frequencies. Each fork produces V beats per second with the previous one. The last is an octave of the first. The fifth fork has a frequency of 90 Hz. Find V and the frequency of the first and the last tuning forks.

Answer»

Data : n_{i + 1} – n_i = Y, n₁₂ = 2n₁ , n₅= 90 Hz n₂ – n₁ = Y beats/s

∴ n₂ = n₁ + Y beats/s

Similarly, n₃ = n₂ + Y = n₁+ Y + Y

∴ n₃ = n₁ + 2Y = n₁ + (3 – 1) Y

∴ n_x = n₁ + (x – 1) Y

Similarly, n₁₂ = n₁ + (12 – 1) Y = n₁ + 11Y

∴ n₁₂ = 2n₁ = n₁ + 11Y

∴ n₁ = 11Y

Also, n₅ = n₁ + (5 – 1) Y = n₁ + 4Y

∴ n₅ = 11Y + 4Y = 15Y

∵ n₅ = 90 Hz

∴ 15Y = 90

∴ Y = 6

∴ n₁ = 11Y beats/s = 11 × 6 beats/s = 66 Hz

and n₁₂ = 2n₁ = 2 (66) = 132 Hz