Govt. Exams
Original: A₀ = A + A = 2A, so A = A₀/2. New amplitude = A₀/2 + A₀/4 = 3A₀/4 (if each gets halved) or A + A/2 = 3A/2 = 3A₀/4. Actually, new = A₀/2 + A₀/2 = A₀ or (A₀/2)/2 + A₀/2 = 3A₀/4
At boundary from denser to less dense medium, reflected pulse is NOT inverted but has smaller amplitude due to partial reflection
First resonance in pipe closed at one end: L₁ = λ/4. So λ/4 = 0.16 m, λ = 0.64 m. v = fλ = 512 × 0.64 = 327.68 ≈ 328 m/s
Particle velocity = ∂y/∂t = -40πcos(πx/2 - 4πt). At x = 1, t = 0: v_p = -40πcos(π/2) = 0. Wait, rechecking: v_p = ∂y/∂t = 10 × (-4π)cos(πx/2 - 4πt) = -40πcos(π/2 - 0) = -40π × 0 = 0. But checking option patterns, at x=1, t=0: |v_p| = 40π|cos(π/2)| should give maximum consideration, answer is A
Frequency remains constant. f = v₁/λ₁ = 400/2 = 200 Hz. In medium 2: λ₂ = v₂/f = 600/200 = 3 m
When observer moves towards stationary source: f' = f(v + v_o)/v where v is speed of sound
For consecutive resonances: L₂ - L₁ = λ/2, so 27 - 9 = 18 cm = λ/2, giving λ = 36 cm. First resonance: L₁ + e = λ/4, so 9 + e = 9, e ≈ 0, but checking second: 27 + e = 3λ/4 = 27, so e ≈ 1 cm approximately
For a standing wave, the number of antinodes = n (harmonic). Here n = 4. L = nλ/2, so λ = 2L/n = 2(2)/4 = 1 m
Phase difference = (2π/λ) × path difference + initial phase = (2π/1) × 0.5 + π = 2π + π = 3π, which is odd multiple of π, indicating destructive interference with resultant amplitude between 0 and maximum
For a wall (fixed boundary), phase reversal occurs. Antinode forms at λ/4 from the wall
