Oscillations and Waves: NEET PYQ, 2018–2026

Answer: (A)

The speed of a wave is the product of its frequency and wavelength, v = νλ. So λ = v/ν = 350 m s⁻¹ / 500 Hz = 0.7 m.

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)

For a source moving away from a stationary observer, the wavelength heard is lengthened by the distance the source travels during one period. The observed wavelength is lambda' = (v + vs)/f. With v = 340 m/s, vs = 20 m/s, and f = 500 Hz, the result is 360/500 = 0.72 m.

Answer: (B)

A pipe closed at one end has a node at the closed end and an antinode at the open end, so it supports only the odd harmonics: 250, 750, 1250, 1750 Hz and so on. 500 Hz, an even harmonic, is missing.

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)

The principle of superposition states that where waves overlap, the resultant displacement at each point is the algebraic sum of the individual displacements. The overlap is temporary. After the waves pass through each other, each one continues as if the other had never been there, with its shape, amplitude and speed unchanged in a uniform medium. Pulses do not interact permanently or exchange energy during the crossing.

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: (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: (B)

For a mass m on a spring of force constant k, the angular frequency is ω = √(k/m) and the period is T = 2π√(m/k). The period depends on the square root of the mass. If the mass is multiplied by 4, the period is multiplied by √4 = 2. The original period is 1.0 s, so the new period is 2.0 s. Check: T ∝ √m, so the ratio of periods is √(4m/m) = 2.