Oscillations NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Oscillations question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Oscillations previous year questions
24 questions. Tap an option to check your answer.
Q1NEET 2021
A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is :
Not quite — the answer is D.
k = F/x = 10/0.05 = 200 N/m. T = 2π√(m/k) = 2π√(2/200) = 2π × 0.1 = 0.628 s. 6.28 s would result from forgetting to convert 5 cm into metres.
Q2NEET 2020 · Q145
The phase difference between displacement and acceleration of a particle in a simple harmonic motion is :
Not quite — the answer is A.
In SHM a = −ω²x, so acceleration is always opposite to displacement, a phase difference of π rad. The π/2 difference is between displacement and velocity.
Q3NEET 2019 · Q10
The displacement of a particle executing simple harmonic motion is given by y = A₀ + A sinωt + B cosωt. Then the amplitude of its oscillation is given by:
Not quite — the answer is B.
A₀ only shifts the mean position; A sinωt + B cosωt combine into a single sinusoid of amplitude √(A² + B²). Adding A₀ to it gives the maximum displacement from the origin, not the amplitude about the mean position.
Q4NEET 2019 · Q19
Average velocity of a particle executing SHM in one complete vibration is:
Not quite — the answer is D.
After one complete vibration the particle returns to its starting point, so displacement is zero and average velocity is zero. Aω is the maximum speed, not the average velocity.
Q5NEET 2018 · Q25
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s² at a distance of 5 m from the mean position. The time period of oscillation is
Not quite — the answer is B.
In SHM a = ω²x, so ω² = 20/5 = 4 and ω = 2 rad/s. T = 2π/ω = π s; 2π s would correspond to ω = 1 rad/s.
Q6NEET 2017 · Q147
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:
Not quite — the answer is C.
Speed v = ω√(A² − x²) = ω√(9 − 4) = √5 ω and acceleration a = ω²x = 2ω² (in cm units). Setting them equal gives ω = √5/2 rad/s, so T = 2π/ω = 4π/√5 s. √5/2π is the frequency, not the period.
Q7NEET 2017 · Q166
A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k′. Then they are connected in parallel and force constant is k″. Then k′ : k″ is:
Not quite — the answer is C.
Spring constant is inversely proportional to length, so the pieces (l/6, 2l/6, 3l/6) have constants 6k, 3k and 2k. In series they rebuild the original spring, k′ = k; in parallel k″ = 6k + 3k + 2k = 11k. Hence k′ : k″ = 1 : 11.
Q8NEET 2016 Phase 2 · Q20
A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is
Not quite — the answer is D.
T ∝ √m, so 3/5 = √(m/(m + 1)) and 9/25 = m/(m + 1). Then 9m + 9 = 25m, giving m = 9/16 kg. Inverting the ratio leads to 16/9 kg.
Q9AIPMT 2015
A particle is executing SHM along a straight line. Its velocities at distances x₁ and x₂ from the mean position are V₁ and V₂, respectively. Its time period is:
Not quite — the answer is C.
V₁² = ω²(a² − x₁²) and V₂² = ω²(a² − x₂²). Subtracting, V₁² − V₂² = ω²(x₂² − x₁²), so ω = √((V₁² − V₂²)/(x₂² − x₁²)) and T = 2π√((x₂² − x₁²)/(V₁² − V₂²)). The inverted ratio gives 1/T, not T.
Q10AIPMT 2015
When two displacements represented by y₁ = a sin(ωt) and y₂ = b cos(ωt) are superimposed the motion is:
Not quite — the answer is D.
a sin ωt + b cos ωt = √(a² + b²) sin(ωt + φ) with tan φ = b/a: a single SHM of the same ω. The two are 90° out of phase, so amplitudes add as √(a² + b²), not (a + b)/2.
Q11AIPMT 2011 Mains
Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is
Not quite — the answer is D.
x = A sin φ = A/2 gives φ = π/6 for one particle and 5π/6 for the other, since they move in opposite directions at that displacement. Phase difference = 5π/6 − π/6 = 2π/3. Zero would mean they move together, and π would mean they cross only at the mean position.
Q12AIPMT 2011 Prelims
Out of the following functions representing motion of a particle which represents SHM? (A) y = sin ωt − cos ωt (B) y = sin³ωt (C) y = 5 cos(3π/4 − 3ωt) (D) y = 1 + ωt + ω²t²
Not quite — the answer is D.
sin ωt − cos ωt = √2 sin(ωt − π/4) and 5 cos(3π/4 − 3ωt) are single sinusoids, so both are SHM. sin³ωt = (3 sin ωt − sin 3ωt)/4 is periodic but a mix of two frequencies, so not SHM, and 1 + ωt + ω²t² is not even periodic. So only the first and third functions are SHM; the claim that only the last function fails is wrong because sin³ωt also fails.
Q13AIPMT 2010 Prelims · Q18
The displacement of a particle along the x-axis is given by x = a sin²ωt. The motion of the particle corresponds to
Not quite — the answer is B.
x = a sin²ωt = (a/2)(1 − cos 2ωt). This is a cosine oscillation of amplitude a/2 about the shifted mean position x = a/2 with angular frequency 2ω, so d²(x − a/2)/dt² = −4ω²(x − a/2): it is SHM of frequency 2ω/2π = ω/π. Calling it non-SHM ignores the shift of the mean position; ω/2π uses the wrong angular frequency.
Q14AIPMT 2010 Prelims · Q19
The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be
Not quite — the answer is A.
T = 2π√(M/k), so T ∝ √M. Doubling the mass gives T′ = √2 T. 2T would need four times the mass.
Q15AIPMT 2009 Prelims · Q5
A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be:
Not quite — the answer is C.
v = ω√(a² − x²) with ω = 2π/T. At x = a/2, v = (2π/T)√(3a²/4) = (2π/T)(√3a/2) = πa√3/T. The value with 2T in the denominator forgets that ω already carries the factor 2π.
Q16AIPMT 2009 Prelims · Q32
Which one of the following equations of motion represents simple harmonic motion? Where k, k₀, k₁ and a are all positive.
Not quite — the answer is A.
Put X = x + a; then acceleration = −kX, which is SHM about x = −a with ω² = k. Acceleration = +kx or k(x + a) points away from equilibrium, and the x² term makes the motion non-linear.
Q17AIPMT 2008 Prelims · Q17
Two simple harmonic motions of angular frequency 100 and 1000 rad s⁻¹ have the same displacement amplitude. The ratio of their maximum accelerations is
Not quite — the answer is C.
Maximum acceleration = ω²A. For equal amplitudes the ratio is (100/1000)² = 1 : 10². 1 : 10 is the ratio of maximum velocities (ωA), not accelerations.
Q18AIPMT 2008 Prelims · Q19
A point performs simple harmonic oscillation of period T and the equation of motion is given by x = a sin(ωt + π/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
Not quite — the answer is A.
v = aω cos(ωt + π/6). Setting v = aω/2 gives ωt + π/6 = π/3, so ωt = π/6 and t = (π/6)(T/2π) = T/12. T/6 would be the answer if the initial phase π/6 were ignored.
Q19AIPMT 2007 Prelims · Q9
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is
Not quite — the answer is D.
If x = A sin ωt, then v = Aω cos ωt and a = −Aω² sin ωt, so acceleration leads velocity by π/2 = 0.5π. A phase difference of π exists between displacement and acceleration, not velocity and acceleration.
Q20AIPMT 2007 Prelims · Q27
The particle executing simple harmonic motion has a kinetic energy K₀ cos² ωt. The maximum values of the potential energy and the total energy are respectively
Not quite — the answer is C.
Kinetic energy is maximum (K₀) at the mean position, where all energy is kinetic, so total energy = K₀. Potential energy is maximum at the extremes and also equals K₀. 2K₀ wrongly adds the two maxima.
Q21AIPMT 2007 Prelims · Q37
A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
Not quite — the answer is B.
x = a sin ωt; x = a/2 when ωt = π/6, so t = (π/6)/(2π/T) = T/12. T/4 is the time to reach the full amplitude.
Q22AIPMT 2006 Prelims · Q3
A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then :
Not quite — the answer is B.
When the block is pushed down by y, the extra buoyant force is Aρgy, so it behaves like a spring of constant k = Aρg. Hence T = 2π√(m/(Aρg)), which varies as 1/√A (and as 1/√ρ, √m). T is not proportional to √ρ or 1/ρ, and it increases, not decreases, with m.
Q23AIPMT 2005 Prelims · Q17
The circular motion of a particle with constant speed is :
Not quite — the answer is D.
Uniform circular motion repeats after every revolution, so it is periodic. It is not SHM, because SHM is motion along a line under a restoring force ∝ displacement; only the projection of circular motion is SHM.
Q24AIPMT 2005 Prelims · Q18
A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is :
Not quite — the answer is D.
v_max = Aω, so ω = 31.4/5 = 6.28 rad/s and f = ω/2π = 1 Hz. Taking 6.28 as the frequency or dividing wrongly gives 2 Hz or 0.5 Hz.
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