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Oscillations is a chapter you can finish in a week and keep for life. Every question in NEET(UG) comes back to one equation, a = −ω2x, and to the handful of time periods that follow from it. Where students lose marks is in the energy part — forgetting that kinetic and potential energy trade places twice in every single swing. Ten questions below. Give them ten honest minutes.
What this chapter carries in NEET(UG)
- Usually 2 to 3 questions every year, which is 8 to 12 marks.
- The simple pendulum and the spring–mass system are asked every year.
- Energy in SHM and the phase relations come up almost as often.
What the syllabus says
In the syllabus (Unit 10, Oscillations and Waves): Periodic motion — time period, frequency, displacement as a function of time, periodic functions. Simple harmonic motion and its equation, phase, oscillations of a spring with restoring force and force constant, energy in SHM as kinetic and potential, and the simple pendulum with the derivation of its time period.
Taken out in 2024: free, forced and damped oscillations; resonance.
Taken from the NMC syllabus for NEET (UG) 2026, the latest one published. See the full syllabus, unit by unit.
Formulas you must know by heart
| Displacement in SHM | x = A sin(ωt + φ) |
| Velocity | v = ω√(A2 − x2), maximum ωA |
| Acceleration | a = −ω2x, maximum ω2A |
| Time period | T = 2π/ω |
| Spring and mass | T = 2π√(m/k) |
| Simple pendulum | T = 2π√(L/g) |
| Kinetic energy | KE = ½mω2(A2 − x2) |
| Potential energy | PE = ½mω2x2 |
| Total energy | E = ½mω2A2 |
| Springs in series | 1/k = 1/k1 + 1/k2 |
| Springs in parallel | k = k1 + k2 |
Look at the last three energy lines together. Kinetic and potential energy each rise and fall twice in one full oscillation, but their sum never changes. That single sentence answers a large share of the energy questions asked in this chapter.
Ten questions. Give yourself ten minutes.
Write your answers down before you scroll. A question you looked up is a question you have not answered.
Q1. If the length of a simple pendulum is made four times, its time period becomes
- (a) double
- (b) half
- (c) four times
- (d) unchanged
Q2. In simple harmonic motion the acceleration is largest at the
- (a) mean position
- (b) extreme position
- (c) point midway between them
- (d) same at every point
Q3. A particle performs SHM of amplitude A. Its kinetic energy equals its potential energy at a displacement of
- (a) A/2
- (b) A/√2
- (c) A
- (d) A/4
Q4. If the mass hanging from a spring is made four times, the time period of oscillation becomes
- (a) double
- (b) half
- (c) four times
- (d) unchanged
Q5. The phase difference between the displacement and the acceleration of a particle in SHM is
- (a) 0
- (b) π/2
- (c) π
- (d) 2π
Q6. A simple pendulum is hung in a lift that moves downward with an acceleration a. Its time period
- (a) increases
- (b) decreases
- (c) stays the same
- (d) becomes zero
Q7. The total energy of a particle in SHM is proportional to
- (a) A
- (b) A2
- (c) 1/A
- (d) A3
Q8. The displacement of a particle is y = 5 sin(10t) metre, with t in seconds. Its time period is about
- (a) 0.314 s
- (b) 0.628 s
- (c) 10 s
- (d) 1 s
Q9. In SHM the speed of the particle is largest at the
- (a) extreme position
- (b) mean position
- (c) displacement A/2
- (d) it is the same everywhere
Q10. Two springs of force constants k1 and k2 are joined end to end. The equivalent force constant is
- (a) k1 + k2
- (b) k1k2/(k1 + k2)
- (c) (k1 + k2)/2
- (d) k1 − k2
Answer key, with a hint for each
| Q1 | (a) | T ∝ √L, and √4 = 2. |
| Q2 | (b) | a = −ω2x, so the acceleration grows with displacement and is largest where x is largest. |
| Q3 | (b) | Put ½mω2(A2 − x2) = ½mω2x2, which gives 2x2 = A2. |
| Q4 | (a) | T ∝ √m. Four times the mass, twice the period. |
| Q5 | (c) | The minus sign in a = −ω2x means the acceleration always points opposite to the displacement, which is a phase difference of π. |
| Q6 | (a) | The effective gravity becomes g − a, which is smaller, and T ∝ 1/√g. A smaller g means a longer period. |
| Q7 | (b) | E = ½mω2A2. Double the amplitude and the energy goes up four times. |
| Q8 | (b) | ω = 10 rad/s, so T = 2π/10 ≈ 0.628 s. |
| Q9 | (b) | v = ω√(A2 − x2) is largest when x = 0. |
| Q10 | (b) | Joined end to end means in series, so the reciprocals add. |
Now mark yourself honestly
NEET(UG) marks the way it marks. Use the same rule here:
- +4 for a right answer
- −1 for a wrong one
- 0 for a blank
So ten questions here are out of 40. In the real paper, 45 Physics questions are out of 180.
If you scored 32 or more, the basics of this chapter are secure. Move to harder questions, not more of these.
If you scored below 24, do not feel bad about it. You have just found out something useful before the exam did. Go back to the formula table above, learn it properly, and sit these ten again tomorrow.
And if you guessed a few and they landed, count those as wrong. A guess that works once will not work in May.
The full chapter, on a clock
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