Formula
Time period of spring
For mass m attached to spring of force constant k: T = 2π √(m/k). Independent of amplitude. ω = √(k/m).
-- NCERT Class 11 Physics, Ch. 13, p. 272Spring Oscillations
8 MCQs6 revision cards9-step worked example
Source: NCERT Oscillations and WavesPYQ coverage: NEET 2021, 2025Official key: NTA-verifiedLast updated: 25 Sep 2026
Spring Oscillations, explained for NEET
A block on a horizontal frictionless surface is attached to a spring. You pull it, release it, and it oscillates. The period of that oscillation is the single formula this lesson owns: T = 2π√(m/k), where m is the mass of the block and k is the spring constant (NCERT Class 11 Physics, Chapter 13 — Oscillations, page 268).
The trap that costs marks: students believe that changing the amplitude changes the period. It does not. The formula contains m and k — no A anywhere. If a NEET question says "the amplitude is doubled, find the new period," the answer is: the period is unchanged. This is a high-frequency distractor in spring-oscillation problems.
Why is period independent of amplitude? At larger amplitude the block travels a longer path, but it also moves faster (the restoring force is stronger at larger displacement). These two effects cancel exactly, keeping the time for one complete cycle the same.
What k means physically. The spring constant k = F/x (Hooke's law). A stiffer spring (larger k) pulls the block back harder, so the oscillation is faster and T is smaller. A heavier block (larger m) has more inertia, so the oscillation is slower and T is larger. The square root means that quadrupling the mass only doubles the period.
Watch-out for NEET: when a problem gives you the force needed to stretch a spring by a certain length, extract k first (k = F/x), then substitute into T = 2π√(m/k). Forgetting the 2π factor or inserting amplitude into the period formula are both common distractor traps.
Can you answer these Spring Oscillations MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The period of oscillation of a mass-spring system on a frictionless horizontal surface depends on:
Show answer and why every option is right or wrong
Answer: C. T = 2π√(m/k) contains only mass m and spring constant k. Period is independent of amplitude for ideal SHM (NCERT Class 11 Physics, Chapter 13, page 268).
Why A is wrong: A is wrong because amplitude does not appear in T = 2π√(m/k). Period is independent of amplitude in ideal SHM (trap: amplitude-dependence claim).
Why B is wrong: B is wrong because amplitude does not affect the period, and mass is a factor — not amplitude (trap: amplitude-dependence claim).
Why D is wrong: D is wrong because amplitude has no role in determining the period of an ideal mass-spring oscillator (trap: amplitude-dependence claim).
MCQ 2Easy RecallPractice
In the formula T = 2π√(m/k) for a mass-spring oscillator, k represents:
Show answer and why every option is right or wrong
Answer: D. k is the spring constant defined by Hooke's law: F = −kx, so k = F/x, the restoring force per unit displacement (NCERT Class 11 Physics, Chapter 13, page 268).
Why A is wrong: A is wrong because k is not an energy quantity. It is defined as force per unit displacement (F/x), not energy per unit displacement.
Why B is wrong: B is wrong because force per unit velocity describes a damping coefficient (b in F = −bv), not a spring constant. k relates force to displacement, not velocity.
Why C is wrong: C is wrong because k has units N/m (force per length), not J/kg (energy per mass). It characterizes the spring stiffness, not energy.
MCQ 3Easy RecallPractice
If the amplitude of a mass-spring oscillator on a frictionless surface is doubled, the period:
Show answer and why every option is right or wrong
Answer: D. T = 2π√(m/k) has no amplitude term. For ideal SHM, period is independent of amplitude (NCERT Class 11 Physics, Chapter 13, page 268).
Why A is wrong: A is wrong because T does not depend on amplitude. Doubling A increases the path length but also increases the average speed proportionally, leaving T unchanged (trap: amplitude-dependence claim).
Why B is wrong: B is wrong because √2 scaling would require T ∝ √A, but no such dependence exists in the ideal spring formula (trap: amplitude-dependence claim).
Why C is wrong: C is wrong because halving would require T ∝ 1/A, which contradicts T = 2π√(m/k) where amplitude is absent (trap: amplitude-dependence claim).
MCQ 4Direct ApplicationPractice
A spring stretches by 0.10 m when a force of 5.0 N is applied. The spring constant k is:
Show answer and why every option is right or wrong
Answer: A. k = F/x = 5.0 N / 0.10 m = 50 N/m (NCERT Class 11 Physics, Chapter 13, page 268).
Why B is wrong: B is wrong because 5.0 N/m uses only the force value. k = F/x requires dividing 5.0 N by 0.10 m, giving 50 N/m.
Why C is wrong: C is wrong because 0.50 N/m results from computing x/F = 0.10/5.0 instead of F/x. The spring constant is force divided by extension, not the inverse.
Why D is wrong: D is wrong because 500 N/m results from a decimal error (dividing by 0.01 m instead of 0.10 m).
MCQ 5Direct ApplicationPractice
A block of mass 4.0 kg is attached to a spring of constant 100 N/m on a frictionless surface. The period of oscillation is:
Show answer and why every option is right or wrong
Answer: B. B is correct. T = 2π√(m/k) = 2π√(4.0/100) = 2π√(0.040) = 2π × 0.20 = 0.40π s. Written as a fraction that is 2π/5 s ≈ 1.26 s. Note that 0.20 = 1/5, so 2π × 0.20 is 2π/5 — not 4π/5 (NCERT Class 11 Physics, Chapter 13, page 268).
Why A is wrong: A is wrong because π/5 s results from dropping the factor of 2 in 2π√(m/k), giving π√(0.040) = π/5 (trap: forgetting the 2π factor).
Why C is wrong: C is wrong because 4π/5 s is exactly twice the correct period. It comes from carrying 2π × 0.20 to the fraction incorrectly: 0.20 is 1/5, so the product is 2π/5. Check it numerically — 4π/5 ≈ 2.51 s, while 2π√0.04 ≈ 1.26 s.
Why D is wrong: D is wrong because 2π/25 s results from using k/m instead of m/k under the square root, giving √(100/4) = 5, then inverting incorrectly.
MCQ 6Direct ApplicationPractice
A mass-spring system has period T. If the mass is replaced by one four times heavier while keeping the same spring, the new period is:
Show answer and why every option is right or wrong
Answer: B. T_new = 2π√(4m/k) = 2 × 2π√(m/k) = 2T. Since T ∝ √m, quadrupling the mass doubles the period (NCERT Class 11 Physics, Chapter 13, page 268).
Why A is wrong: A is wrong because T/2 would apply if period were inversely proportional to mass, but T ∝ √m — increasing mass increases the period, not decreases it.
Why C is wrong: C is wrong because period does depend on mass. T ∝ √m, so changing mass changes the period. Only amplitude changes leave T unchanged.
Why D is wrong: D is wrong because 4T would require T ∝ m (linear), but the actual relation is T ∝ √m. Quadrupling m gives √4 = 2 times the original period, not 4 times.
MCQ 7CalculationPYQ Pattern
A spring requires a force of 2.0 N to stretch it by 0.050 m. A block of mass 0.50 kg is attached to this spring on a frictionless surface and set into oscillation. The period of oscillation is closest to:
Show answer and why every option is right or wrong
Answer: A. A is correct. Step 1: k = F/x = 2.0/0.050 = 40 N/m. Step 2: T = 2π√(m/k) = 2π√(0.50/40) = 2π√(1/80) = 2π/(4√5) = π/(2√5) = π√5/10 s ≈ 0.70 s. Rationalising π/(2√5) multiplies top and bottom by √5, giving π√5/10 — note that is √5, not √10 (NCERT Class 11 Physics, Chapter 13, page 268).
Why B is wrong: B is wrong because π/√10 ≈ 0.99 s. This results from using k = 20 N/m (dividing force by twice the extension or a unit error) instead of the correct k = 40 N/m.
Why C is wrong: C is wrong because π√5/20 s ≈ 0.35 s is half the correct period: it writes T = π√(m/k), dropping the factor 2 from 2π.
Why D is wrong: D is wrong because π√5 s ≈ 7.0 s comes from leaving the extension in centimetres, k = 2.0/5.0 = 0.40 N/m: 2π√(0.50/0.40) = 2π(√5/2) = π√5 s.
MCQ 8CalculationPractice
Two identical springs, each of constant k, are connected in parallel (side by side) to a block of mass m. The period of oscillation compared to a single-spring system is:
Show answer and why every option is right or wrong
Answer: B. Springs in parallel: effective constant k_eff = k + k = 2k. T_parallel = 2π√(m/2k) = (1/√2) × 2π√(m/k) = T_single/√2. The stiffer effective spring gives a shorter period (NCERT Class 11 Physics, Chapter 13, page 268).
Why A is wrong: A is wrong because adding a second spring in parallel doubles the effective stiffness (k_eff = 2k). Since T ∝ 1/√k, the period decreases — it does not stay the same.
Why C is wrong: C is wrong because √2 × T would imply a softer effective spring (k_eff = k/2), which describes springs in series, not parallel. Parallel combination stiffens the system.
Why D is wrong: D is wrong because T/2 would require k_eff = 4k (T ∝ 1/√k, so halving T needs quadrupling k). Two parallel springs give only 2k, producing T/√2.
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Spring Oscillations: quick recall before you leave
How do you solve a Spring Oscillations question? A worked example
- 1
Given
A block of mass 0.50 kg hangs at rest from a vertical spring and stretches it by 4.0 cm. The block is then pulled down a little and released so that it oscillates vertically. Take g = 10 m/s².
- 2
Required
The period of the vertical oscillation.
- 3
Concept
A hanging block settles where the spring force balances its weight, so the static extension x₀ already measures the spring: kx₀ = mg. Once it is set oscillating about that new equilibrium, gravity is entirely accounted for by the shift of the equilibrium point, and the motion about it is the ordinary mass–spring SHM with T = 2π√(m/k) (NCERT Class 11 Physics, Chapter 13, page 268).
Substituting k = mg/x₀ into that period makes the mass cancel, leaving T = 2π√(x₀/g). The static stretch alone gives the period — which is why this arrangement is a quick way to measure g. - 4
Formula
• Static balance: k = mg / x₀• Period: T = 2π√(m/k), equivalently T = 2π√(x₀/g)
- 5
Substitution
• k = (0.50 × 10) / 0.040• T = 2π√(0.50 / k)• Cross-check: T = 2π√(0.040 / 10)
- 6
Calculation
• k = 5.0 / 0.040 = 125 N/m• m/k = 0.50 / 125 = 4.0 × 10⁻³ s²• √(4.0 × 10⁻³) = 0.0632 s• T = 2π × 0.0632 = 0.397 s
Cross-check by the mass-free form: x₀/g = 0.040/10 = 4.0 × 10⁻³ s², the same number, so T = 0.397 s again. ✓
Note on exact values: 2π is a mathematical constant and g is problem-defined; the two significant figures in the mass and the extension set the precision. - 7
Final answer
T ≈ 0.40 s
The two routes agreeing is not a coincidence — it is the mass cancelling. Hang a different block on the same spring and the extension changes in exactly the proportion that keeps T fixed. - 8
Common trap
The frequent error is adding gravity to the restoring force and expecting a different period from the horizontal case. It is already included: hanging the spring moves the equilibrium down by x₀ and changes nothing else, so the period is the same as it would be for that spring and block lying on a frictionless table. The second trap is using the extension as an amplitude — 4.0 cm is where the block RESTS, not how far it swings, and the amplitude does not enter T at all.
- 9
Similar NEET-style question
A spring stretches 2.5 cm under a hanging load and oscillates with period T. What extension would give a period of 2T? Approach: T ∝ √x₀, so doubling T needs four times the extension: 10 cm. The load required to produce it is four times as large, but the load never appears in the answer.
What to remember before solving Spring Oscillations questions
Which Spring Oscillations formulas do you need for NEET?
1 formula — click to collapse
Period of mass-spring oscillator
Period of horizontal spring with mass m, spring constant k. Independent of amplitude.
| Symbol | Quantity | SI Unit |
|---|---|---|
| T | period | s |
| m | mass | kg |
| k | spring constant | N/m |
Valid when
- Hooke's law spring
- No damping
- Small enough amplitude to stay in elastic regime
Where do students lose marks on Spring Oscillations?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Overthinking
Student claims SHM period depends on amplitude. For ideal SHM (Hooke's law spring or simple pendulum at small angle), period is INDEPENDENT of amplitude.
When it triggers
Question gives changes in amplitude and asks for new period.
How to avoid
T = 2π√(m/k) (spring) or 2π√(L/g) (pendulum, small angle) — neither depends on A. Only at large pendulum angles does T pick up a small amplitude correction.
Root cause: concept gap
Correction
Ideal SHM: T = 2π√(m/k) (spring) or 2π√(L/g) (pendulum, small angle) — no amplitude dependence. Doubling amplitude does not change period.
More in Oscillations and Waves: 4 exam traps and mistakes · 9 formulas · 5 question patterns from its other lessons.
Spring Oscillations questions from past NEET papers
2 questions from NEET 2021, 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
1√(k₁/k₂)
2k₂/k₁
3k₁/k₂
4√(k₂/k₁)
NTA Answer: Option 1(final)
NEET 2021
10.628 s
20.0628 s
36.28 s
43.14 s
NTA Answer: Option 1(final)
How does NEET ask about Spring Oscillations?
1 recurring pattern from past papers — click to collapse
Given spring extension/compression with given force, find k, then find period when given mass m. T = 2*pi*sqrt(m/k).
Multi StepEasy
Common distractors
forgets 2pi factor
Drops 2*pi
uses amplitude in period
Believes T depends on amplitude
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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