Why this chapter pays for itself twice
Oscillations is a short chapter with an unusually high return, because its mathematics reappears elsewhere almost unchanged. Waves is built directly on it - a wave is SHM propagated through space. In Alternating Current, the equations governing an LC circuit are identical in form to a spring-mass system, with inductance playing the role of mass and the reciprocal of capacitance playing the role of the spring constant.
That means the effort spent here is not confined here. A student who genuinely understands why ω = √(k/m) will recognise ω = 1/√(LC) as the same statement, which is a substantial saving later.
How JEE actually asks Oscillations
NTA publishes no chapter-wise weightage, so treat online figures as coaching estimates from past papers, and check the official NTA information bulletin for the syllabus and pattern applying to your session.
The observable pattern is that direct period calculations are the minority. More common are questions that give a slightly unusual arrangement - springs in combination, a vertical spring, a pendulum in an accelerating lift, a body floating in a liquid - and ask whether the motion is simple harmonic and what its period is. Energy questions, particularly the fraction of total energy that is kinetic at a given displacement, are also a reliable presence.
Key concepts, compressed
- ✓SHM is defined by a = −ω²x. If you can show the restoring acceleration is proportional to displacement and opposite in sign, the motion is simple harmonic and ω is the square root of the constant of proportionality.
- ✓Angular frequency ω relates to period and frequency by ω = 2π/T = 2πf. Mixing ω and f is the most common arithmetic error in the chapter.
- ✓Velocity in SHM is v = ω√(A² − x²), which is maximum ωA at the mean position and zero at the extremes.
- ✓Acceleration is a = −ω²x, so it is zero at the mean position and maximum ω²A at the extremes - exactly opposite to velocity.
- ✓Total mechanical energy is ½mω²A², constant throughout the motion and proportional to the square of the amplitude.
- ✓Damping reduces amplitude over time; resonance occurs when a driving frequency matches the natural frequency, and damping limits how large the resonant amplitude becomes.
Formulas you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Displacement | x = A sin(ωt + φ) | φ is the phase constant |
| Velocity | v = ω√(A² − x²) | maximum ωA at x = 0 |
| Acceleration | a = −ω²x | maximum ω²A at x = ±A |
| Angular frequency | ω = 2π/T = 2πf | |
| Spring-mass period | T = 2π√(m/k) | independent of g |
| Simple pendulum period | T = 2π√(L/g) | small angles only |
| Springs in series | 1/k = 1/k₁ + 1/k₂ | softer than either |
| Springs in parallel | k = k₁ + k₂ | stiffer than either |
| Total energy | E = ½mω²A² = ½kA² | |
| Kinetic energy at x | KE = ½mω²(A² − x²) | |
| Potential energy at x | PE = ½mω²x² |
The five mistakes that cost the most marks
- ✓Assuming any back-and-forth motion is simple harmonic. It qualifies only when the restoring force is proportional to displacement - which is why a pendulum is SHM at small angles but not at large ones.
- ✓Confusing angular frequency ω with frequency f. They differ by a factor of 2π, so substituting one for the other changes the answer by about 6.28.
- ✓Thinking a spring-mass system's period depends on gravity. It does not - T = 2π√(m/k) holds horizontally, vertically and on the Moon. Only the equilibrium position shifts.
- ✓Assuming acceleration is zero wherever velocity is zero. At the extremes velocity is zero and acceleration is at its maximum - the exact opposite of the mean position.
- ✓Reversing series and parallel for springs. Springs in parallel add stiffness directly; in series they combine reciprocally and give a softer effective spring, which is the opposite of resistors.
Practice set 1: SHM basics
Questions 1 to 3 all refer to the same motion, described by x = 5 sin(4t) with x in metres and t in seconds.
1. For x = 5 sin(4t), find the amplitude, angular frequency and time period.
Amplitude 5 m, angular frequency 4 rad/s, period about 1.57 s. Comparing with the standard form x = A sin(ωt + φ) gives A = 5 and ω = 4 directly. Then T = 2π/ω = 2π/4 ≈ 1.57 s. The coefficient of t is ω, not f - reading it as a frequency in hertz is the standard error here.
2. For the same motion, what is the maximum velocity?
20 m/s. Maximum velocity is ωA = 4 × 5 = 20 m/s, occurring at the mean position where x = 0. This follows from v = ω√(A² − x²), which is largest when x is zero. Notice that the maximum speed depends on both amplitude and frequency, so a faster oscillation of the same amplitude is faster at the centre.
3. For the same motion, what is the maximum acceleration?
80 m/s². Maximum acceleration is ω²A = 16 × 5 = 80 m/s², occurring at the extremes where the displacement is largest. Compare with the previous answer: velocity peaks at the centre and acceleration peaks at the edges, and the two never peak together.
4. What is the defining condition for simple harmonic motion?
a = −ω²x: the acceleration must be proportional to the displacement from the mean position and directed opposite to it. Every SHM result follows from this one relation. When a question describes an unfamiliar oscillating arrangement, the intended method is to derive the restoring force, show it is proportional to displacement, and read ω² off the constant of proportionality.
5. At the mean position of an SHM, what are the velocity and acceleration?
Velocity is maximum at ωA; acceleration is zero. At x = 0 the restoring force vanishes, so there is nothing accelerating the body at that instant - yet it is moving fastest, because all the energy is kinetic there. This is the same relationship as a pendulum bob at the bottom of its swing.
6. At the extreme position, what are the velocity and acceleration?
Velocity is zero; acceleration is maximum at ω²A, directed back toward the mean position. The body is momentarily at rest but experiencing the largest restoring force, which is what turns it around. Answering that acceleration is also zero because the body is at rest is the classic error, and it is the same mistake as the ball at the top of its flight in Kinematics.
7. For x = 5 sin(4t), what is the velocity when the displacement is 3 m?
16 m/s. Using v = ω√(A² − x²) = 4 × √(25 − 9) = 4 × √16 = 4 × 4 = 16 m/s. A useful check: this is less than the maximum 20 m/s, as it must be, and would fall to zero at x = 5. The relation shows velocity depends on position, not on time, which is often the faster route.
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Practice set 2: springs and pendulums
8. A 2 kg mass hangs from a spring of force constant 200 N/m. What is the period of its oscillation?
About 0.63 s. T = 2π√(m/k) = 2π√(2/200) = 2π√0.01 = 2π × 0.1 ≈ 0.628 s. Note that no information about gravity was needed, and none was used - the period of a spring-mass system is the same whether it hangs vertically or slides horizontally.
9. What is the frequency of that oscillation?
About 1.59 Hz. f = 1/T = 1/0.628 ≈ 1.59 Hz. Equivalently f = ω/2π with ω = √(k/m) = √100 = 10 rad/s, giving f = 10/6.283 ≈ 1.59 Hz. Checking a frequency both ways is a quick guard against the ω-versus-f confusion.
10. What is the period of a simple pendulum of length 1 m?
About 1.99 s. T = 2π√(L/g) = 2π√(1/10) = 2π × 0.3162 ≈ 1.99 s. This is close to 2 s, which is why a pendulum of about one metre is the traditional seconds pendulum - it takes roughly one second for each swing in either direction.
11. The length of a pendulum is quadrupled. What happens to its period?
It doubles. T is proportional to √L, so multiplying the length by four multiplies the period by two. To halve a pendulum's period you would need to reduce its length to a quarter. The mass of the bob has no effect at all, which surprises students who expect a heavier bob to swing differently.
12. Does the period of a spring-mass system change if it is taken to the Moon?
No. T = 2π√(m/k) contains no g, so the period is unchanged. What does change is the equilibrium extension of a vertically hung spring, which is mg/k and therefore smaller on the Moon. A pendulum, by contrast, would swing much more slowly there because T = 2π√(L/g) does contain g.
13. Two springs of 200 N/m each are connected in series. What is the effective spring constant?
100 N/m. For springs in series, 1/k = 1/200 + 1/200 = 2/200, so k = 100 N/m. Springs in series are softer than either individual spring, because the same force stretches both and the extensions add. This is the opposite of resistors in series, which is where the confusion comes from.
14. The same two springs are connected in parallel. What is the effective spring constant?
400 N/m. In parallel the constants add: k = 200 + 200 = 400 N/m. Both springs stretch by the same amount and share the load, so the combination is stiffer. Since T = 2π√(m/k), the parallel combination oscillates twice as fast as the series one for the same mass.
Practice set 3: energy in SHM
Questions 15 to 17 refer to a 2 kg mass performing SHM with ω = 4 rad/s and amplitude 5 m.
15. What is the total mechanical energy of the oscillation?
400 J. E = ½mω²A² = ½ × 2 × 16 × 25 = 400 J. This total stays constant throughout the motion; only its division between kinetic and potential changes. Note the energy is proportional to the square of the amplitude, so doubling the amplitude quadruples the energy required.
16. What fraction of the total energy is kinetic when the displacement is half the amplitude?
Three quarters, so 75%. KE = ½mω²(A² − x²), and with x = A/2 this becomes ½mω²(A² − A²/4) = (3/4) × ½mω²A² = 0.75E. For our numbers that is 300 J. The result depends only on the ratio x/A, not on the mass, frequency or amplitude individually.
17. What fraction is potential at that same point?
One quarter, so 25% - which is 100 J here. PE = ½mω²x² = ½mω²(A²/4) = 0.25E. It must be the remainder, since the two always sum to the total. Note the asymmetry: at half the amplitude the energy is still three-quarters kinetic, because energy depends on the square of displacement.
18. At what displacement are the kinetic and potential energies equal?
At x = A/√2, about 0.707 of the amplitude. Setting ½mω²(A² − x²) = ½mω²x² gives A² − x² = x², so x² = A²/2 and x = A/√2. Students often guess A/2, which is where the split is 75:25 rather than 50:50 - question 16 exists precisely to make that contrast visible.
19. At what frequency do the kinetic and potential energies oscillate, compared with the displacement?
At twice the frequency. Both energies depend on the square of a sinusoid, and squaring a sine doubles its frequency. So a body completing one oscillation passes through maximum kinetic energy twice - once moving each way through the centre. This is a frequently asked detail and cannot be reasoned out without the squaring argument.
20. How does the total energy of an oscillation depend on amplitude?
It is proportional to the square of the amplitude, since E = ½mω²A². Tripling the amplitude therefore requires nine times the energy. This is also why damped oscillations lose amplitude quickly at first - the energy loss per cycle is roughly proportional to the energy present, and energy falls faster than amplitude does.
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Practice set 4: phase, damping and resonance
21. What is the phase difference between displacement and velocity in SHM?
π/2, or 90°, with velocity leading. If x = A sin(ωt) then v = Aω cos(ωt) = Aω sin(ωt + π/2). Physically, velocity peaks a quarter of a cycle before displacement does - which is another way of saying velocity is greatest at the centre while displacement is greatest at the edges.
22. What is the phase difference between displacement and acceleration?
π, or 180° - they are exactly out of phase. This follows directly from a = −ω²x: the minus sign means acceleration always points opposite to displacement. When the body is at its positive extreme, its acceleration is at its most negative, driving it back toward the centre.
23. How does the amplitude of a damped oscillation change with time?
It decays exponentially, following A(t) = A₀e^(−bt/2m) for light damping. The period changes only slightly for weak damping, which is why a swinging pendulum keeps nearly constant timing while visibly losing height. The energy, going as amplitude squared, decays twice as fast in the exponent.
24. What is resonance?
The large-amplitude response that occurs when the frequency of a periodic driving force matches the system's natural frequency. At resonance, energy is fed in at exactly the right moment in each cycle, so it accumulates rather than cancelling. It is why pushing a swing in time with its motion works and pushing at random does not.
25. What effect does damping have on resonance?
It reduces the peak amplitude and broadens the resonance curve, and shifts the peak slightly below the undamped natural frequency. With no damping at all the amplitude at resonance would grow without limit, which is why real structures are deliberately damped - the damping is what makes the response finite.
Practice set 5: combinations and applied SHM
26. Two SHMs of the same frequency, with amplitudes 3 cm and 4 cm, act along the same line with a phase difference of 90°. What is the resultant amplitude?
5 cm. Amplitudes combine like vectors with the phase difference as the angle between them, so with a 90° difference the resultant is √(3² + 4²) = 5 cm. Had they been in phase the answer would be 7 cm, and exactly out of phase it would be 1 cm - so the phase difference matters as much as the amplitudes.
27. A 2 kg mass hangs from a spring of constant 200 N/m. By how much does the spring stretch at equilibrium?
0.1 m. At equilibrium the spring force balances the weight: kx₀ = mg, so x₀ = mg/k = 2 × 10/200 = 0.1 m. This defines the new mean position about which the mass oscillates - gravity shifts where the centre is, but as the next question shows, it does not affect the period.
28. What is the period of oscillation of that hanging mass?
About 0.63 s - identical to the horizontal case in question 8. Measuring displacement from the new equilibrium position, the weight is exactly cancelled by the initial spring extension, and the net restoring force is again −kx. Gravity relocates the mean position and then drops out of the dynamics entirely.
29. A simple pendulum of length 1 m hangs in a lift accelerating upward at 10 m/s². What is its period?
About 1.40 s. In the accelerating frame the effective gravity is g + a = 20 m/s², so T = 2π√(L/g(eff)) = 2π√(1/20) = 2π × 0.2236 ≈ 1.40 s. The pendulum swings faster than the 1.99 s of question 10, because the restoring force has effectively been strengthened.
30. What happens to a pendulum in a freely falling lift?
It stops oscillating. In free fall the effective gravity is g − g = 0, so the restoring force vanishes and the period T = 2π√(L/g(eff)) becomes infinite. The bob simply floats in whatever position it was left. This is the limiting case of question 29 and the pendulum equivalent of the weightless astronaut in Gravitation.
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How to study this chapter efficiently
- ✓For any unfamiliar oscillating system, derive the restoring force first and check whether it is proportional to displacement. If it is, ω² is the constant of proportionality and everything else follows.
- ✓Write ω = 2πf on every page of working until the distinction is automatic. This single confusion accounts for more wrong answers in this chapter than any conceptual gap.
- ✓Learn the two period formulas as a contrasting pair: the spring has no g, the pendulum does. Questions exploit that difference constantly.
- ✓For energy questions, work in fractions of A rather than absolute numbers. The kinetic fraction at x = A/2 is 3/4 regardless of mass, amplitude or frequency.
- ✓Use the effective-gravity idea for any pendulum in an accelerating frame or an electric field - replace g with g(eff) and the standard formula still applies.
- ✓Study this chapter immediately before Waves. The two share almost all their mathematics, and doing them together roughly halves the work.
Turn this into active practice
The recurring failure in this chapter is not computational. Students who can derive a period correctly still lose marks on questions asking where velocity is maximum, at what displacement the energies are equal, or whether a described motion is simple harmonic at all - because those are recognition questions, and recognition only develops by being tested.
The JEE Oscillations quiz on QUFF generates fresh questions across SHM basics, spring and pendulum systems, energy and resonance, marks them instantly and explains each answer. Do a mixed set rather than one topic at a time, since the value here lies in identifying what kind of question you are looking at.
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