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. 272Restoring Force Constant
8 MCQs5 revision cards9-step worked example
Source: NCERT Oscillations and WavesPYQ coverage: NEET 2026Official key: NTA-verifiedLast updated: 26 Sep 2026
Restoring Force Constant, explained for NEET
A block on a spring is pulled aside and released. It oscillates. Why? Because the spring exerts a restoring force — a force that always points back toward the equilibrium position. This force is what makes simple harmonic motion possible.
For an ideal spring obeying Hooke's law, the restoring force is F = −kx, where k is the force constant (also called spring constant) and x is displacement from equilibrium. The negative sign encodes the "restoring" nature: displace right, force pulls left; displace left, force pushes right. The force constant k has units N/m and measures stiffness — a larger k means a stiffer spring and a stronger pull back to equilibrium (NCERT Class 11 Physics, Chapter 13, page 267).
Connect this to the period of oscillation: T = 2π√(m/k). A stiffer spring (larger k) gives a shorter period — the system snaps back faster. A heavier mass (larger m) gives a longer period — more inertia to overcome.
The trap that costs marks: students believe that changing the amplitude changes the period. It does not. The formula T = 2π√(m/k) contains neither amplitude A nor the initial displacement. Double the amplitude and the block travels farther, but it also moves faster (larger restoring force at larger displacement). These effects cancel exactly, leaving the period unchanged. This is a high-frequency confusion in NEET — when you see a problem that changes amplitude and asks for the new period, the answer is: the period stays the same.
Watch out: the force constant k is a property of the spring, not the mass. If a problem says "a 2.0 N/cm spring," convert to SI (200 N/m) before substituting into T = 2π√(m/k).
Can you answer these Restoring Force Constant 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 SI unit of force constant (spring constant) is:
Show answer and why every option is right or wrong
Answer: D. D is correct. Force constant k = F/x, so its unit is newton per metre (N/m). See NCERT Class 11 Physics, Chapter 13, page 267.
Why A is wrong: A is wrong because N·m is the unit of torque (or work), not of force per unit displacement.
Why B is wrong: B is wrong because N·m² has dimensions of force × area, unrelated to the ratio F/x.
Why C is wrong: C is wrong because kg/s has dimensions of mass/time, which does not match force/displacement (MLT⁻²/L = MT⁻²).
MCQ 2Easy RecallPractice
In Hooke's law F = −kx, the negative sign indicates that the force:
Show answer and why every option is right or wrong
Answer: C. C is correct. The negative sign means the force opposes the displacement — it is always directed back toward the equilibrium position. This is the defining feature of a restoring force. See NCERT Class 11 Physics, Chapter 13, page 267.
Why A is wrong: A is wrong because the negative sign means the force acts opposite to displacement, not in the same direction.
Why B is wrong: B is wrong because the magnitude of the force |F| = kx actually increases with displacement; the negative sign indicates direction, not magnitude reduction.
Why D is wrong: D is wrong because Hooke's law F = −kx involves k and x only; mass does not appear in this relation.
MCQ 3Easy RecallPractice
A spring has a force constant of 400 N/m. This means that:
Show answer and why every option is right or wrong
Answer: C. C is correct. By definition, k = F/x, so k = 400 N/m means 400 N of force produces 1 m of extension (within the elastic limit). See NCERT Class 11 Physics, Chapter 13, page 267.
Why A is wrong: A is wrong because the energy stored at 1 m extension is E = ½kx² = ½ × 400 × 1² = 200 J, not 400 J.
Why B is wrong: B is wrong because the force constant says nothing about the natural (unstretched) length of the spring.
Why D is wrong: D is wrong because k defines stiffness (force per unit extension), not a maximum load capacity.
MCQ 4Direct ApplicationPractice
A spring stretches by 5.0 cm when a force of 10 N is applied. The force constant of the spring is:
Show answer and why every option is right or wrong
Answer: B. B is correct. k = F/x = 10 N / 0.050 m = 200 N/m. Note the unit conversion from cm to m before substitution.
Why A is wrong: A is wrong — this results from using x = 0.20 m (misreading 5.0 cm as 20 cm) or a different arithmetic error.
Why C is wrong: C is wrong — this results from dividing 10 N by 0.10 m, i.e., doubling the extension by mistake.
Why D is wrong: D is wrong — this results from using x = 0.02 m (misreading 5.0 cm as 2.0 cm).
MCQ 5Direct ApplicationPractice
A body of mass 0.50 kg is attached to a spring of force constant 200 N/m 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. T = 2π√(m/k) = 2π√(0.50/200) = 2π√(2.5 × 10⁻³) = 2π × 0.050 = 0.314 s ≈ 0.31 s. See NCERT Class 11 Physics, Chapter 13, page 267.
Why B is wrong: B is wrong — 0.10 s is 2√(m/k): the 2π factor has lost its π (trap: forgetting 2π factor).
Why C is wrong: C is wrong — 0.050 s is √(m/k) alone, with the 2π factor dropped completely.
Why D is wrong: D is wrong — this is far too large; it corresponds to m/k ≈ 0.025, not 0.0025, suggesting a unit error such as using m = 5.0 kg.
MCQ 6Direct ApplicationPractice
A mass-spring system oscillates with amplitude A and period T. If the amplitude is doubled to 2A while everything else remains unchanged, the new period is:
Show answer and why every option is right or wrong
Answer: A. A is correct. T = 2π√(m/k) — the period depends only on mass and spring constant, not on amplitude. Doubling amplitude does not change the period. This is a documented high-frequency confusion (trap: period-amplitude dependence claim).
Why B is wrong: B is wrong — halving the period would require quadrupling k or reducing m to one-quarter; amplitude change does not affect T.
Why C is wrong: C is wrong — this assumes T ∝ A, i.e., that period scales linearly with amplitude. Amplitude does not appear in T = 2π√(m/k).
Why D is wrong: D is wrong — this assumes T ∝ A², an even stronger (and incorrect) amplitude dependence.
MCQ 7CalculationPYQ Pattern
A spring extends by 4.0 cm when a force of 8.0 N is applied. A block of mass 0.80 kg is now attached to this spring and set into oscillation. The period of oscillation is closest to:
Show answer and why every option is right or wrong
Answer: B. B is correct. Step 1: k = F/x = 8.0 N / 0.040 m = 200 N/m. Step 2: T = 2π√(m/k) = 2π√(0.80/200) = 2π√(4.0 × 10⁻³) = 2π × 0.0632 = 0.397 s ≈ 0.40 s.
Why A is wrong: A is wrong — 0.20 s is half the period: the time from one extreme to the other, not a full oscillation back to the start.
Why C is wrong: C is wrong — 4.0 s comes from leaving the extension in centimetres, k = 8.0/4.0 = 2.0 N/cm used as 2.0 N/m: 2π√(0.80/2.0) ≈ 4.0 s.
Why D is wrong: D is wrong — 0.063 s is √(m/k) alone, with the 2π factor dropped.
MCQ 8CalculationPractice
Two springs have force constants k₁ = 100 N/m and k₂ = 300 N/m. They are connected in parallel and a 1.0 kg block is attached. The period of oscillation of the block is closest to:
Show answer and why every option is right or wrong
Answer: D. D is correct. For parallel springs, k_eff = k₁ + k₂ = 100 + 300 = 400 N/m. T = 2π√(m/k_eff) = 2π√(1.0/400) = 2π × 0.050 = 0.314 s ≈ 0.31 s.
Why A is wrong: A is wrong — 0.73 s comes from the series combination, 1/k = 1/k₁ + 1/k₂, giving k = 75 N/m; springs side by side in parallel add, k = k₁ + k₂ (trap: confusing series vs parallel spring combination).
Why B is wrong: B is wrong — 0.63 s comes from using only k₁ = 100 N/m, ignoring the stiffer spring: T = 2π√(1.0/100) ≈ 0.63 s.
Why C is wrong: C is wrong — 0.36 s comes from using only k₂ = 300 N/m, ignoring the second spring: T = 2π√(1.0/300) ≈ 0.36 s.
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Restoring Force Constant: quick recall before you leave
How do you solve a Restoring Force Constant question? A worked example
- 1
Given
• Mass: m = 0.20 kg• Equation of motion: a = −64x (SI units)• Displacement of interest: x = 5.0 cm = 0.050 m
- 2
Required
(a) Force constant k. (b) Restoring force at x = 0.050 m. (c) Period T.
- 3
Concept
The force constant does not have to come from a measured stretch. Newton's second law applied to a spring gives ma = −kx, so a = −(k/m)x. Comparing that with the defining form of SHM, a = −ω²x, identifies ω² = k/m — the coefficient sitting in front of x in the equation of motion IS k/m (NCERT Class 11 Physics, Chapter 13, page 268).
So a stated a–x relation already contains the spring constant; it only needs the mass to extract it. This is the route NEET uses when no extension is given. - 4
Formula
• Compare a = −64x with a = −ω²x → ω² = 64• k = mω²• Restoring force: F = −kx• Period: T = 2π/ω
- 5
Substitution
ω² = 64 s⁻², so ω = 8.0 rad/s
k = 0.20 × 64
F = −(k) × 0.050
T = 2π / 8.0 - 6
Calculation
k = 0.20 × 64 = 12.8 N/m
F = −12.8 × 0.050 = −0.64 N
T = 2π / 8.0 = 0.785 s
Note on exact values: the 64 is given exactly by the problem's equation and the 2π is a mathematical constant; the two significant figures in the mass set the precision of k and F. - 7
Final answer
(a) k = 12.8 N/m (b) F = −0.64 N, the minus sign meaning it points back towards the mean position, opposite to the displacement (c) T = 0.785 s ≈ 0.79 s
Notice that the period never needed k at all: T = 2π/ω follows straight from ω = 8.0 rad/s. The force constant was required only for the FORCE. - 8
Common trap
Two errors are common here. The first is reading 64 as ω rather than ω², giving k = 0.20 × 4096 and a period eight times too short; the coefficient of x is ω², because the defining relation is a = −ω²x. The second is reporting F = +0.64 N. The minus sign is the entire content of Hooke's law — it is what makes the force RESTORING, and dropping it describes a spring that pushes the block further out, which would never oscillate at all.
- 9
Similar NEET-style question
A 0.50 kg body executes SHM with a = −100x (SI units). Find the force constant and the frequency. Approach: ω² = 100, so ω = 10 rad/s and k = mω² = 0.50 × 100 = 50 N/m. Then f = ω/2π = 10/(2π) = 1.6 Hz.
What to remember before solving Restoring Force Constant questions
Which Restoring Force Constant 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
More in Oscillations and Waves: 6 exam traps and mistakes · 9 formulas · 5 question patterns from its other lessons.
Restoring Force Constant questions from past NEET papers
1 question from NEET 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
14
22
31/2
41/4
NTA Answer: Option 4(final)
How does NEET ask about Restoring Force Constant?
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
NCERT refs: Class 11 Physics Chapter 13, p.267
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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