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Oscillations rewards a student who can read an equation. Given x = A sin(ωt + φ), you should be able to write down the amplitude, the maximum speed, the maximum acceleration and the time period without pausing. JEE Main then builds on that with cut springs, springs in series and parallel, and the share between kinetic and potential energy. Ten questions below. Give them ten honest minutes.
What this chapter carries in JEE Main
- Usually 1 to 2 questions every year, which is 4 to 8 marks.
- Reading quantities straight off the SHM equation is the standard question.
- Spring combinations and the energy of SHM fill the rest.
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 NTA syllabus for JEE Main 2027. See the full syllabus, unit by unit.
Formulas you must know by heart
| Displacement | x = A sin(ωt + φ) |
| Maximum speed and acceleration | ωA and ω2A |
| Velocity at displacement x | v = ω√(A2 − x2) |
| Spring and mass | T = 2π√(m/k) |
| Simple pendulum | T = 2π√(L/g) |
| Spring cut into n equal parts | each piece has force constant nk |
| Springs in series and parallel | 1/k = 1/k1 + 1/k2, and k = k1 + k2 |
| Energies | KE = ½mω2(A2 − x2), PE = ½mω2x2 |
| Total energy | E = ½mω2A2 |
| Two SHMs combined | A = √(A12 + A22 + 2A1A2cosφ) |
The line about cutting a spring catches people out. A shorter spring is a stiffer spring, so cutting one in half does not halve the force constant — it doubles it, and the time period of a mass on it falls by a factor of √2.
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. A particle moves as x = 4 sin(2t + π/6) metre, with t in seconds. Its maximum speed is
- (a) 4 m/s
- (b) 8 m/s
- (c) 2 m/s
- (d) 16 m/s
Q2. For the same particle, the maximum acceleration is
- (a) 8 m/s2
- (b) 16 m/s2
- (c) 4 m/s2
- (d) 32 m/s2
Q3. A spring of force constant k is cut into two equal halves. The force constant of each half is
- (a) k/2
- (b) 2k
- (c) k
- (d) 4k
Q4. A simple pendulum is taken to the moon, where gravity is one sixth of that on earth. Its time period becomes
- (a) 6 times larger
- (b) √6 times larger
- (c) one sixth
- (d) 1/√6 times
Q5. Two simple harmonic motions of the same frequency, with amplitudes 3 cm and 4 cm and a phase difference of π/2, are added. The resultant amplitude is
- (a) 7 cm
- (b) 5 cm
- (c) 1 cm
- (d) 3.5 cm
Q6. A particle performs SHM of amplitude A. At what displacement from the mean position is its kinetic energy equal to its potential energy?
- (a) A/4
- (b) A/2
- (c) A/√2
- (d) A
Q7. Two springs of force constants k₁ and k₂ are joined end to end and a mass m hangs from the pair. The time period of its vertical oscillation is
- (a) 2π√(m(k₁+k₂)/k₁k₂)
- (b) 2π√(m/(k₁+k₂))
- (c) 2π√(mk₁k₂/(k₁+k₂))
- (d) 2π√((k₁+k₂)/m)
Q8. The potential energy of a particle performing SHM is greatest at the
- (a) mean position
- (b) extreme position
- (c) displacement A/2
- (d) same everywhere
Q9. A 1 kg mass oscillates on a spring of force constant 100 N/m. Its frequency is about
- (a) 1.59 Hz
- (b) 10 Hz
- (c) 0.1 Hz
- (d) 100 Hz
Q10. A particle in SHM of time period T starts from the mean position. The time it takes to reach half the amplitude is
- (a) T/4
- (b) T/8
- (c) T/12
- (d) T/6
Answer key, with a hint for each
| Q1 | (b) | ω = 2 and A = 4, so the maximum speed is ωA = 8. |
| Q2 | (b) | ω2A = 4 × 4. |
| Q3 | (b) | A shorter spring stretches less for the same force, so it is stiffer. Half the length means twice the constant. |
| Q4 | (b) | T ∝ 1/√g, so a sixth of g makes the period √6 times longer. |
| Q5 | (b) | With a phase difference of π/2 the amplitudes add like perpendicular vectors: √(9 + 16). |
| Q6 | (c) | KE = ½mω²(A² − x²) and PE = ½mω²x². Setting them equal gives A² − x² = x², so x = A/√2. At that one point each of them carries half the total energy. |
| Q7 | (a) | Springs in series add their compliances: 1/k = 1/k₁ + 1/k₂, so k = k₁k₂/(k₁+k₂). Put that into T = 2π√(m/k). Joining springs in series makes the pair softer, so the period gets longer. |
| Q8 | (b) | PE = ½mω2x2, which is greatest where x is greatest. |
| Q9 | (a) | ω = √(100/1) = 10 rad/s, so f = 10/2π. |
| Q10 | (c) | x = A sin(2πt/T). Put x = A/2, so 2πt/T = π/6. |
Now mark yourself honestly
JEE Main 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, 25 Physics questions are out of 100.
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 January.
The full chapter, on a clock
Ten questions tell you a little. Twenty-five tell you the truth.
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