Govt. Exams
For a tube closed at one end (water surface), first resonance occurs at L = λ/4. Given f = 512 Hz, v = 340 m/s, λ = v/f = 340/512 ≈ 0.664 m. So λ/4 ≈ 0.166 m = 16.6 cm. The effective length includes end correction (≈1 cm for open end), so L_eff = 16.2 + 1 = 17.2 cm, which equals λ/4 within measurement precision.
Intensity ∝ 1/r². I₂/I₁ = (r₁/r₂)² = (1/10)² = 1/100. I₂ = 10⁻⁸/100 = 10⁻¹⁰ W/m²
Phase difference = (2π/λ) × path difference = (2π/20) × 10 = π radian
A closed organ pipe resonates at odd multiples: f₁, 3f₁, 5f₁... The next resonance after 200 Hz is 3×200 = 600 Hz
From y = A cos(kx) sin(ωt), we have k = π, so λ = 2π/k = 2π/π = 2 cm
Using Snell's law: n₁sin(θ₁) = n₂sin(θ₂). 1.5 × sin(30°) = 1 × sin(θ₂). θ₂ = 48.75°
From y = A sin(2πx/λ - 2πt/T), we get λ = 4 m and T = 0.5 s. v = λ/T = 4/0.5 = 8 m/s
Diffraction is more prominent when wavelength is comparable to or larger than the aperture size
f₁ = (1/2L)√(T/μ) where μ = 0.01/1 = 0.01 kg/m. f₁ = 0.5 × √(40/0.01) = 31.6 Hz
Using Doppler formula: f' = f × v/(v + v_source) = 600 × 340/(340 + 30) = 547 Hz
