Restoring Force Constant

8 MCQs2 revision cards9-step worked example

Source: NCERT Oscillations and WavesPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 5).

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).

Practice 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:

MCQ 2Easy RecallPractice

In Hooke's law F = −kx, the negative sign indicates that the force:

MCQ 3Easy RecallPractice

A spring has a force constant of 400 N/m. This means that:

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:

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:

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:

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:

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:

Quick recall before you leave

Worked Example

1
Given
- Applied force: F = 10.0 N - Compression: x = 5.0 cm = 5.0 × 10⁻² m - Mass of block: m = 2.0 kg
2
Required
Period of oscillation T.
3
Concept
The spring obeys Hooke's law (F = kx), so we first find k from the static load data. Then we use T = 2π√(m/k) to find the period of SHM.
4
Formula
k = F/x (from Hooke's law) T = 2π√(m/k) (period of mass-spring oscillator)
5
Substitution
k = 10.0 N / (5.0 × 10⁻² m) T = 2π√(2.0 kg / k)
6
Calculation
k = 10.0 / 0.050 = 200 N/m m/k = 2.0 / 200 = 1.0 × 10⁻² s² √(1.0 × 10⁻²) = 0.10 s T = 2π × 0.10 = 0.6283 s **Exact constants note:** The factors 2 and π are exact mathematical constants. They do not limit significant figures. The answer's precision is governed by the given data (2 significant figures).
7
Final answer
T ≈ 0.63 s (Rounded to 2 significant figures, matching the precision of the given data.)
8
Common trap
The amplitude of oscillation (5.0 cm initial compression) does not appear in the period formula. If the problem changes the amplitude — say the spring is now compressed by 10 cm — the period remains 0.63 s. This is the amplitude-independence trap documented in trap: period amplitude dependence: students who include amplitude in their period calculation get a wrong answer.
9
Similar NEET-style question
A spring of force constant 800 N/m supports a 0.50 kg mass. The mass is pulled down 3.0 cm and released. What is the time period? (Answer: T = 2π√(0.50/800) = 2π × 0.025 = 0.157 s ≈ 0.16 s. The 3.0 cm pull-down is irrelevant to the period.) ---

Before solving, remember these

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, p. 5

Formulas

10 formulas — click to collapse

Beat frequency

f_{beat} = ∣ f_{1} - f_{2} ∣

When two waves of nearly equal frequencies superpose, amplitude oscillates at the difference frequency.

Symbol	Quantity	SI Unit
f_beat	beat frequency	Hz
f1, f2	superposed frequencies	Hz

Valid when

Linear superposition
f1, f2 close in value

Period of simple pendulum (small angle)

T = 2 π \frac{L}{g}

Period of simple pendulum of length L. Holds for small amplitudes (sin theta ~ theta).

Symbol	Quantity	SI Unit
T	period	s
L	pendulum length	m
g	gravity	m/s^2

Valid when

Small angular amplitude (typically <15°)
Massless string
Point bob

SHM displacement

x (t) = A cos (ω t + ϕ); ω = 2 π f

Displacement in simple harmonic motion. Velocity = -A*omega*sin(omega*t+phi); a = -omega^2 * x.

Symbol	Quantity	SI Unit
A	amplitude	m
omega	angular frequency	rad/s
phi	phase	rad
T	period	s
f	frequency	Hz

Valid when

Restoring force linear (F = -kx)
No damping

Total energy in SHM

E = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2}

Total mechanical energy is constant. Oscillates between KE (max at x=0) and PE (max at x=±A).

Symbol	Quantity	SI Unit
E	total energy	J
k	spring constant	N/m
A	amplitude	m
m	mass	kg
omega	angular frequency	rad/s

Valid when

Conservative SHM (no damping)
Elastic regime

Period of mass-spring oscillator

T = 2 π \frac{m}{k}

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

Standing wave in closed-end pipe

f_{n} = \frac{( 2 n - 1 ) v}{4 L}

Pipe closed at one end has only odd harmonics: f, 3f, 5f, ...

Symbol	Quantity	SI Unit
f_n	n-th harmonic	Hz
v	sound speed	m/s
L	pipe length	m

Valid when

Closed at one end (open at other)
End correction neglected

Standing wave in open-open pipe

f_{n} = \frac{n v}{2 L}

Pipe open at both ends has all harmonics. Same formula as string.

Symbol	Quantity	SI Unit
f_n	n-th harmonic	Hz
v	sound speed	m/s
L	pipe length	m

Valid when

Open at both ends
End correction neglected

Standing wave frequencies on fixed-fixed string

f_{n} = \frac{n v}{2 L}

Allowed frequencies on string fixed at both ends. n=1 fundamental; harmonics 2f, 3f, ...

Symbol	Quantity	SI Unit
f_n	n-th harmonic	Hz
v	wave speed on string	m/s
L	string length	m
n	harmonic number	-

Valid when

String fixed at both ends
Wave speed v as defined above

Speed of sound in gas (Newton-Laplace)

v = \frac{γ P}{ρ}

Speed of sound in gas. Adiabatic index gamma, pressure P, density rho. Increases with sqrt(T).

Symbol	Quantity	SI Unit
v	speed of sound	m/s
gamma	adiabatic index	-
P	pressure	Pa
rho	density	kg/m^3

Valid when

Ideal gas
Adiabatic compression/expansion of sound waves

Wave speed on string

v = \frac{T}{μ}

Speed of transverse wave on string under tension T, linear mass density mu.

Symbol	Quantity	SI Unit
v	wave speed	m/s
T	tension	N
mu	linear mass density	kg/m

Valid when

Stretched uniform string
Small amplitude

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

6 items — click to collapse

Trap

Pendulum: mass-dependence claim

Category: Overthinking

Student writes T as depending on bob mass. Simple pendulum T = 2π√(L/g); independent of m.

When it triggers

Question changes pendulum bob mass and asks for new period.

How to avoid

Mass cancels in derivation (gravitational mass = inertial mass). Mass changes the bob's KE and PE proportionally; period unaffected.

Trap

SHM period: amplitude dependence claim

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.

Trap

Pipe harmonics: open vs closed end

Category: Similar Terms

Student includes even harmonics in a closed-end pipe. Closed pipe has only ODD harmonics (f, 3f, 5f, ...).

When it triggers

Question describes pipe closed at one end (e.g. resonance tube).

How to avoid

Open both ends: all harmonics, f_n = nv/(2L). Closed one end: odd only, f_n = (2n-1)v/(4L). Fundamental of closed pipe is HALF that of open pipe of same L.

Mistake

Includes bob mass in simple pendulum period formula.

Root cause: concept gap

Correction

Simple pendulum T = 2π√(L/g) — independent of mass. Equivalence of inertial and gravitational mass cancels m.

Mistake

Claims SHM period depends on amplitude.

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.

Mistake

Includes even harmonics in a closed-end pipe.

Root cause: concept gap

Correction

Closed-end pipe has only ODD harmonics (f, 3f, 5f, ...). Open-both-ends pipe has all (f, 2f, 3f, ...). Reason: closed end has displacement node and pressure antinode.

Past Year Questions

11 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

A pipe open at both ends has a fundamental frequency f in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to: f

12f

22 3f

NTA Answer: Option 3(final)

NEET 2025

Two identical point masses P and Q, suspended from two separate massless springs of spring constants k 1 and k 2, respectively, oscillate vertically. If their maximum speeds are the same, the ratio (AQ/AP) of the amplitude AQ of mass Q to the amplitude AP of mass P is k k

22 k k 2 1

3k 1

4k 2 k k 2 1

NTA Answer: Option 1(final)

NEET 2024Revised key

 π If x =5sinπt + m represents the motion of a particle executing simple harmonic motion, the amplitude  3 and time period of motion, respectively, are

15 cm, 2 s

25 m, 2 s

35 cm, 1 s

45 m, 1 s

NTA Answer: Option 2(revised_final)

NEET 2024Revised key

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half x its original length, then the new time period of oscillation is times its original time period. Then the value of 2 x is:

32 3

NTA Answer: Option 2(revised_final)

NEET 2023

The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is

12 : 1

21 : 3

33 : 1

41 : 2

NTA Answer: Option 1(final)

NEET 2023

The x-t graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at t = 2 s is π2 π2

1– ms–2

2ms–2 8 16 π2 π2

3– ms–2

4ms–2 16 8

NTA Answer: Option 3(final)

NEET 2022

If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is

11 : 2

21 : 1

32 : 1

41 : 2

NTA Answer: Option 4(final)

NEET 2022

Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

211

410

NTA Answer: Option 2(final)

NEET 2021

A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is

10.628 s

20.0628 s

36.28 s

43.14 s

NTA Answer: Option 1(final)

NEET 2020

U U U U A U B Sfl U U U 6 Hz fl 25. In a guitar, two strings A and B made of same flS U U – B fl U U material are slightly out of tune and produce beats , flS fl U of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. 7 Hz – A fl 530 Hz , B If the frequency of A is 530 Hz, the original fl — frequency of B will be :

1537 Hz

2523 Hz

3524 Hz

4536 Hz

NTA Answer: Option 3(final)

NEET 2020

U fl U flS U fl U 29. The phase difference between displacement and U — acceleration of a particle in a simple harmonic motion is :

1zero

2rad 3

3rad 2

4rad 2

NTA Answer: Option 2(final)

How NEET usually asks this

6 recurring patterns from past papers — click to collapse

Two strings/forks slightly out of tune produce beats; find one frequency given the other and beat frequency.

Direct ApplicationEasy

Common distractors

treats beat as sum

Adds frequencies instead of subtracting

Given pendulum length and g, find T. Or given mass change, observe T unchanged.

Direct ApplicationEasy

Common distractors

expects mass dependence

Assumes T depends on bob mass

Open-open pipe (all harmonics) vs closed-end pipe (odd only). Compare frequency ratios.

InterpretationMedium

Common distractors

treats closed pipe like open

Includes even harmonics in closed pipe

Phase between displacement, velocity, acceleration in SHM. v leads x by π/2; a leads x by π.

InterpretationEasy

Common distractors

uses pi 2 instead of pi

Confuses v-x phase with a-x phase

Given spring extension/compression with given force, find k, then find period when given mass m. T = 2pisqrt(m/k).

Multi StepEasy

Common distractors

forgets 2pi factor

Drops 2*pi

uses amplitude in period

Believes T depends on amplitude

Given tension change, find new wave speed on string. v ∝ sqrt(T).

Direct ApplicationEasy

Common distractors

uses linear tension scaling

Treats v ∝ T not sqrt(T)

Sources

NCERT refs: Class 11 Physics Chapter 13, p.5

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Restoring Force Constant

Lesson

Practice MCQs

Quick recall before you leave

Hooke's law (restoring force)

Period of mass-spring oscillator

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources