Oscillations and Waves: JEE Advanced PYQ, 2007–2026

Answer: (D)

In a frame accelerating upward with acceleration a, the effective gravity is g_eff = g + a = 4g/3. Since T = 2π√(L/g_eff), T'/T = √(g/g_eff) = √(3/4) = √3/2. The period decreases because the effective gravity is larger.

Answer: (A)

In a pipe closed at one end, resonance occurs when the effective column length is an odd multiple of λ/4. Adjacent resonances therefore differ by λ/2, because the added half-loop is a node-to-node (or antinode-to-antinode) spacing. The end correction e adds to each measured length equally and cancels in the difference. So λ/2 = 62.0 − 20.0 = 42.0 cm and λ = 84 cm = 0.84 m. Then v = fλ = 400 × 0.84 = 336 m s⁻¹. As a check, 20 + e = 21 cm gives e = 1 cm, and 62 + e = 63 cm = 3 × 21 cm, which is consistent. Using either length alone as an exact λ/4 or 3λ/4 gives wrong speeds.

Answer: (A)

The unknown true frequency f is eliminated by taking the ratio. Approaching: f₁ = fv/(v − v_s). Receding: f₂ = fv/(v + v_s). So f₁/f₂ = (v + v_s)/(v − v_s) = 660/540 = 11/9. Cross-multiplying gives 9v + 9v_s = 11v − 11v_s, so v_s = v/10 = 34 m/s.

Answer: (B)

The two blocks oscillate together with ω² = k/(M + m) = 200/4 = 50 s⁻². The upper block gets its acceleration only from static friction, whose maximum is μmg. So its maximum acceleration without slipping is μg = 5 m s⁻². In SHM the maximum acceleration is ω²A. Setting ω²A ≤ μg gives A ≤ μg/ω² = 5/50 = 0.1 m. In general A_max = μg(M + m)/k = 10 cm.

Answer: (A)

For the moving source, f' = f v/(v − u_s) = f × 340/272 = 1.25 f. For a moving observer and a stationary source, f' = f (v + w)/v. Setting (340 + w)/340 = 1.25 gives w = 0.25 × 340 = 85 m/s. This is larger than 68 m/s because the Doppler shift is not symmetric between source motion and observer motion.

Answer: (D)

Take the positive direction from the source to the observer. Both move in this direction. The observer's motion along the sound's direction lowers the rate of crest arrival, so the numerator is v − v_o = 320. The source's motion towards the observer compresses the wavelength, so the denominator is v − v_s = 300. f' = 1500 × 320/300 = 1600 Hz. The source's closing speed outweighs the observer's retreat, so the net shift is upward.

Answer: (A)

Starting from the mean position, x = A sin(ωt) with ω = 2π/T. Setting x = A/2 gives sin(ωt) = 1/2, so ωt = π/6. Then t = (π/6)/(2π/T) = T/12 = 2.4/12 = 0.2 s. The same result follows from the reference circle: the projection reaches A/2 after the radius vector turns through 30°, which is 1/12 of a full revolution.

Answer: (A)

Intensity depends on the net phase difference, which is the source phase difference plus the path-difference phase (2π/λ)Δx. For a far point on +x, S1 is farther by λ/4, so its wave picks up an extra lag of π/2. S2 itself lags S1 by π/2 at emission, so the two waves arrive with the same phase. The amplitudes add, giving A_res = 2A and I = 4I₀. For a far point on −x, S2 is farther by λ/4. Its wave then lags S1's by π/2 (source lag) + π/2 (path lag) = π, so the waves cancel and I = 0. The result is strongly directional: 4I₀ along +x and 0 along −x.

Answer: (D)

The motion has two halves. In one half the bob swings with the full length L, taking T/2 = 1.0 s. In the other half the string is held by the peg, so the effective length is L − 3L/4 = L/4. That pendulum has period T' = 2π√(L/4g) = T/2 = 1.0 s, so its half-swing takes 0.5 s. The total period is 1.0 + 0.5 = 1.5 s.

Answer: (A)

A standing wave on a string with both ends fixed has the form y = A_max sin(kx) cos(ωt), where the fixed end at x = 0 is a node. For the third harmonic, L = 3λ/2, so λ = 2L/3 = 0.6 m and k = 2π/λ = 10π/3 m⁻¹. At x = 0.1 m, kx = π/3, so the amplitude is 8 mm × sin(π/3) = 8 × 0.866 ≈ 6.9 mm. The wrong options come from using cosine instead of sine, or from using the wrong harmonic's wavelength.