JEE Main 2024 Physics question
Two open organ pipes of length 60 cm and 90 cm resonate at 6th and 5th harmonics respectively. The difference of frequencies for the given modes is Hz. (Velocity of sound in air = 333 m/s)
Answer: 740
Solution
f = nv/2L: f₁ = 6 × 333/(2 × 0.6) = 1665 Hz and f₂ = 5 × 333/(2 × 0.9) = 925 Hz, so Δf = 740 Hz.