PhysicsCBSE Class 11 · Oscillations

Spring-Mass Systems and the Simple Pendulum Class 11

Updated 17 October 2026 · Fact-checked

A mass m on a spring of constant k oscillates with T = 2π√(m/k), independent of g and of orientation. Springs in parallel add (k = k₁ + k₂), springs in series combine as 1/k = 1/k₁ + 1/k₂. A simple pendulum of length L has T = 2π√(L/g_eff), where g_eff is the effective gravity (g ± a in an accelerating lift, zero in free fall); the period is independent of the mass for small angles.

Understand Spring-Mass Systems and the Simple Pendulum

General rule. If the restoring force on a displacement $x$ is $F = - C x$ (with $C$ a constant), the motion is SHM with $T = 2 π m / C$ . Identify $C$ and the inertia $m$ .

Spring-mass system. For a block of mass $m$ on a spring of constant $k$ : $ω = k / m$ and $T = 2 π m / k$ . It holds on a horizontal smooth surface and vertically (gravity only shifts the equilibrium by $x_{0} = m g / k$ ), so $T = 2 π x_{0} / g$ for a vertical spring. At displacement $x$ from the equilibrium the spring force in a vertical oscillation ranges between $k (x_{0} - A)$ and $k (x_{0} + A)$ .

Combinations. Springs in parallel (same extension): $k = k_{1} + k_{2}$ . In series (same force): $\frac{1}{k} = \frac{1}{k _{1}} + \frac{1}{k _{2}}$ . A spring cut into $n$ equal parts has each part of constant $nk$ . Stiffer springs oscillate faster.

Simple pendulum. For small angles, $F \approx - \frac{m g}{L} x$ so
$ $T = 2 π \frac{L}{g} .$ $
It does not depend on the mass or amplitude (small angle). In a lift moving up with acceleration $a$ , $g_{e f f} = g + a$ (a longer weight, shorter period); moving down, $g_{e f f} = g - a$ ; in free fall $g_{e f f} = 0$ and there are no oscillations. A pendulum on a planet of gravity $g$ : $T \propto 1/ g$ . A pendulum of $T = 2$ s is a "seconds pendulum".

Other SHM systems. A liquid column of total length $l$ in a U-tube: $T = 2 π l /2 g$ . A block of mass $m$ and cross-section $A$ floating in a liquid of density $ρ$ : restoring force $- ρ g A y$ , so $T = 2 π m / ρ g A$ .

Platform condition. A block resting on a platform that oscillates vertically with amplitude $A$ stays in contact if $ω^{2} A \leq g$ .

Spring-Mass Systems and the Simple Pendulum: formulas and key facts

Spring-mass: $T = 2 π \frac{m}{k}$; Any orientation.
Parallel / series springs: $k = k_{1} + k_{2}; \frac{1}{k} = \frac{1}{k _{1}} + \frac{1}{k _{2}}$; Combination rules.
Vertical spring: $T = 2 π \frac{x _{0}}{g}, x_{0} = \frac{m g}{k}$; Equilibrium extension.
Simple pendulum: $T = 2 π \frac{L}{g}$; Small angles.
Pendulum in an accelerating lift: $T = 2 π \frac{L}{g \pm a}$; Up: +, down: −.
U-tube liquid: $T = 2 π \frac{l}{2 g}$; Column length l.

How to solve Spring-Mass Systems and the Simple Pendulum questions

Find the restoring force constant C and the moving mass; then T = 2π√(inertia / C).

1Write the net restoring force for a small displacement from equilibrium.
2Identify C (force per unit displacement) and the inertia m.
3Compute T = 2π√(m/C) (or ω = √(C/m)).
4For spring combinations, find k_eff first.
5For a pendulum, find g_eff (include the acceleration of the support).
6Use v_max = Aω and energy ½kA² for the other quantities.

Spring-Mass Systems and the Simple Pendulum in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A car's suspension

Suppose a car of mass 1200 kg on four identical springs sinks 4.9 cm when loaded as a whole (treat as one effective spring).

Estimate its natural frequency of vertical oscillation (g = 9.8 m/s²).

Try it first, then show the solutionSolution

T = 2π√(x₀/g) = 2π√(0.049/9.8) = 2π × 0.0707 = 0.444 s.
f = 1/T = 2.25 Hz.

Answer: About 2.25 Hz.

Memory hook: Only the static compression is needed.

A grandfather clock

Suppose a pendulum clock has a pendulum that takes 1.00 s for one swing (a half period), so T = 2.00 s (g = 9.80 m/s²).

Find its length.

Try it first, then show the solutionSolution

L = gT²/4π² = 9.80 × 4/39.48 = 0.993 m.

Answer: About 0.993 m.

Memory hook: This is the seconds pendulum.

Tricks to remember Spring-Mass Systems and the Simple Pendulum

Short memory aids. Say each one aloud once, then use it on a question the same day.

Mass up, slower
T ∝ √m for a spring; the pendulum ignores m.
Use it when: Mass dependence.
Parallel stiffer, series softer
k adds in parallel, 1/k adds in series.
Use it when: Spring combinations.
Lift changes g
Up-acceleration increases g_eff; free fall makes it zero.
Use it when: Pendulum in a lift.
Static stretch
T = 2π√(x₀/g) for a vertical spring.
Use it when: Hanging spring.

Spring-Mass Systems and the Simple Pendulum: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Spring-mass period · 2 marks · Very short answerBoard level

A 0.30 kg block is attached to a horizontal spring of constant 120 N/m on a smooth table. Find its period and frequency.

Try it first, then show the solutionSolution

T = 2π√(m/k) = 2π√(0.30/120) = 2π × 0.05. (1)
T = 0.314 s; f = 1/T = 3.18 Hz. (1)

Answer: T ≈ 0.31 s; f ≈ 3.2 Hz.

Example 2: Length of a seconds pendulum · 3 marks · Short answerBoard level

Find the period of a simple pendulum of length 0.60 m at g = 9.8 m/s². What length gives a period of exactly 2 s at this place?

Try it first, then show the solutionSolution

T = 2π√(L/g) = 2π√(0.60/9.8) = 1.55 s. (1)
For T = 2 s: L = gT²/4π² = 9.8 × 4/39.48. (1)
L = 0.993 m. (1)

Answer: 1.55 s; 0.993 m.

Example 3: Pendulum in a lift · 3 marks · Short answerHarder board level

A simple pendulum of length 1.6 m has a period T at rest on Earth (g = 10 m/s²). Find T, and the period when the pendulum is in a lift accelerating upwards at 2.5 m/s². What happens in free fall?

Try it first, then show the solutionSolution

T = 2π√(1.6/10) = 2.51 s. (1)
Upward acceleration: g_eff = 10 + 2.5 = 12.5, T = 2π√(1.6/12.5) = 2.25 s. (1)
In free fall g_eff = 0 and the bob has no restoring force, so there are no oscillations (T → ∞). (1)

Answer: 2.51 s; 2.25 s; no oscillation in free fall.

Example 4: Vertical spring · 4 marks · Long answerHarder board level

A 0.50 kg mass hung from a spring stretches it by 5.0 cm in equilibrium (g = 9.8 m/s²). The mass is pulled down a further 3.0 cm and released. Find (a) the spring constant, (b) the period, (c) the maximum speed, (d) the maximum and minimum forces exerted by the spring.

Try it first, then show the solutionSolution

(a) k = mg/x₀ = 0.5 × 9.8/0.05 = 98 N/m. (1)
(b) T = 2π√(m/k) = 2π√(0.5/98) = 0.449 s (same as 2π√(x₀/g)). (1)
(c) ω = 14.0 rad/s; v_max = Aω = 0.03 × 14.0 = 0.42 m/s. (1)
(d) F_max = k(x₀ + A) = 98 × 0.08 = 7.84 N; F_min = k(x₀ − A) = 98 × 0.02 = 1.96 N. (1)

Answer: (a) 98 N/m; (b) 0.449 s; (c) 0.42 m/s; (d) 7.84 N and 1.96 N.

Example 5: Liquid column and floating block · 4 marks · Long answerHOTS / competency

(a) A liquid column of total length 0.40 m in a uniform U-tube is slightly displaced. Show that it executes SHM and find T (g = 10 m/s²). (b) A block of mass 0.20 kg and base area 25 cm² floats in water (ρ = 1000 kg/m³); it is pushed down slightly. Find its period.

Try it first, then show the solutionSolution

(a) If the level in one limb falls by y, the other rises by y; the net unbalanced column is 2y, so restoring force = −2Aρg y while mass = Aρl. (1)
ω² = 2g/l, T = 2π√(l/2g) = 2π√(0.40/20) = 0.89 s. (1)
(b) Extra buoyancy for a depth y: ρgAy, so k_eff = ρgA = 1000 × 10 × 25 × 10⁻⁴ = 25 N/m. (1)
T = 2π√(m/k_eff) = 2π√(0.20/25) = 0.56 s. (1)

Answer: (a) 0.89 s; (b) 0.56 s.

Where students lose marks in Spring-Mass Systems and the Simple Pendulum

The slips examiners see most often, and what to write instead.

Mistake: Taking g into the spring period.
Thinking gravity affects T of a spring.
Do this: T = 2π√(m/k) regardless of g.
Usually costs: The answer mark.
Mistake: Adding spring constants for springs in series.
Mixing up the rules.
Do this: Series: 1/k adds; parallel: k adds.
Usually costs: Method marks.
Mistake: Forgetting the lift's acceleration in a pendulum.
Using g alone.
Do this: g_eff = g ± a.
Usually costs: The answer mark.
Mistake: Using the full length of the cut spring for the constant.
Not scaling k.
Do this: A piece of 1/n length has constant nk.
Usually costs: 1 mark.

Exam tips

Identify the restoring force constant and the inertia in the first line.
Show the series/parallel k calculation separately.
For the lift, write g_eff = g + a (up) or g − a (down).
Quote T and f together if asked for the frequency.
Check that small-angle approximation applies for the pendulum.

Practice questions on Spring-Mass Systems and the Simple Pendulum

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If the length of a simple pendulum is quadrupled and the mass is doubled, its period becomes
(a) 2 times (b) 4 times (c) 2√2 times (d) √2 times
Hint
T ∝ √L and is independent of m.
Answer
(a) 2 times.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A simple pendulum clock stops oscillating inside a freely falling lift.
Reason (R): In free fall the effective acceleration due to gravity for the bob is zero.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
g_eff = 0.
Answer
(a).
Q3 · 3 marks · Short answerHarder board level
Two springs of constants 300 N/m and 600 N/m are connected in series and a 2.0 kg block is attached to the combination. Find the effective spring constant and the period.
Hint
1/k = 1/k₁ + 1/k₂.
Answer
k = (300 × 600)/900 = 200 N/m; T = 2π√(2/200) = 0.63 s.
Q4 · 3 marks · Short answerHarder board level
A spring of constant 30 N/m carries a 0.30 kg mass. The spring is cut into three equal pieces and the mass is attached to one piece. Find the periods before and after.
Hint
Each piece has constant 3k.
Answer
Before: T = 2π√(0.3/30) = 0.628 s. After: k′ = 90 N/m, T′ = 2π√(0.3/90) = 0.363 s = T/√3.
Q5 · 4 marks · Long answerHOTS / competency
A platform oscillates vertically in SHM with amplitude 2.0 cm. (a) Find the greatest frequency at which a block on it stays in contact (g = 10 m/s²). (b) At 2.0 Hz, find the maximum acceleration and the largest and smallest normal force on a 0.50 kg block.
Hint
Contact is lost when the platform's downward acceleration exceeds g.
Answer
(a) ω²A = g gives ω = √(10/0.02) = 22.4 rad/s, f = 3.56 Hz. (b) ω = 4π = 12.57 rad/s, a_max = ω²A = 3.16 m/s². N_max = m(g + a) = 0.5 × 13.16 = 6.58 N (at the lowest point); N_min = m(g − a) = 0.5 × 6.84 = 3.42 N (at the top).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.T of spring-mass?Show answer
2π√(m/k).
2.T of a simple pendulum?Show answer
2π√(L/g).
3.Springs in parallel?Show answer
k₁ + k₂.
4.Vertical spring period?Show answer
2π√(x₀/g).
5.U-tube period?Show answer
2π√(l/2g).
6.Seconds pendulum T?Show answer
2 s.

Spring-Mass Systems and the Simple Pendulum: frequently asked questions

Does the mass affect a pendulum's period?

No; the restoring force and the inertia are both proportional to m, so it cancels.

Why must the angle be small?

sin θ ≈ θ only for small θ; otherwise the motion is not SHM.

What is effective gravity?

The apparent acceleration of free fall in the frame of the pendulum, including the support's acceleration.

Does a vertical spring have a different period from a horizontal one?

No; gravity shifts the equilibrium but not the period.

How is a spring cut in half different?

Each half is stiffer, with twice the spring constant.