JEE Mains Physics · Oscillations

Spring Systems and Restoring Forces

Any restoring force that grows in step with the displacement gives SHM with ω² = (force per unit displacement) ÷ mass; springs are the commonest case, alone, cut or combined.

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Why this matters

Thirty PYQs, twenty-two of them multiple choice, and five from 2026. Eleven cut a spring or join springs to one block; eleven change the mass on a spring or put two masses on one spring; eight find ω for something that is not a spring: a floating block, a tunnel through the earth, a disc on a pivot, a dipole in a field. Every one ends with ω² = restoring force per unit displacement ÷ mass.

Concept 1 of 3: Cut springs and springs in series and in parallel

A spring's constant is inversely proportional to its length: a shorter piece of the same spring is stiffer. Springs one after the other carry the same force and share the stretch, so they are softer together (series). Springs that stretch by the same amount share the force, so they are stiffer together (parallel).

Definition

Cutting: $kl$ is constant. A spring cut in the ratio 1 : 2 gives pieces of $3k$ and $3k/2$ ; cut in half, each half is $2k$ .
Series (same force): $\dfrac{1}{k} = \dfrac{1}{k_1} + \dfrac{1}{k_2}$ . Two equal springs give $k/2$ , so T grows by $\sqrt{2}$ .
Parallel (same stretch): $k = k_1 + k_2$ .
A block between two springs, each fixed to a wall: displace it by x and one spring stretches by x while the other is compressed by x. Both push it back, so this is parallel, $k_1 + k_2$ .
The same holds on an incline: $mg\sin\alpha$ is constant, so it only moves the equilibrium point and does not change ω.
Networks: reduce each chain of springs in series first, then add chains that join the same two points.
Then $T = 2\pi\sqrt{m/k_{eq}}$ .

Cut and combined springs

k \propto \frac{1}{l} \qquad \frac{1}{k_s} = \frac{1}{k_1} + \frac{1}{k_2} \qquad k_p = k_1 + k_2

Worked example

A spring of constant 24 N/m is cut into two pieces whose lengths are in the ratio 1 : 2. Find the constant of each piece. A 0.75 kg block on a smooth floor is then fixed between the two pieces, whose other ends are fixed to walls. Find its period.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q11Moderate

Example 1 · Oscillations · Spring Systems and Restoring Forces

A spring of force constant

15\text{ }N/m

is cut into two pieces. If the ratio of their length is

1:3

, then the force constant of smaller piece is

\_\_\_\_

N/m

A block between two walls is parallel

The springs sit on opposite sides of the block, so they look like a chain. But they stretch and compress by the same x and both push the block back: k = k₁ + k₂, not the series value.

Cutting a spring makes it stiffer

A shorter piece of the same spring has a larger constant. Cut in half, each half is 2k, not k/2.

Gravity along an incline does not change ω

For a block between springs on a smooth incline, mg sin α is a constant force. It shifts the equilibrium point but leaves the restoring force −(k₁ + k₂)x unchanged.

Drill 10 more on cut springs and springs in series and in parallel

Concept 2 of 3: Mass on a spring, and two masses on one spring

A heavier mass on the same spring swings more slowly:

T \propto \sqrt{m}

. When nothing is fixed to a wall and two masses sit on the ends of one spring, both move about their centre of mass, and the right mass to use is the reduced mass.

Definition

$T = 2\pi\sqrt{m/k}$ ; $f \propto 1/\sqrt{m}$ and $f \propto \sqrt{k}$ . Four times the mass doubles T.
A vertical spring has the same T. Gravity only stretches it by $\Delta = mg/k$ , so $T = 2\pi\sqrt{\Delta/g}$ .
Equal masses on springs $k_1$ and $k_2$ : equal amplitudes give $v_{max}$ in the ratio $\sqrt{k_1/k_2}$ ; equal $v_{max}$ give amplitudes in the ratio $\sqrt{k_2/k_1}$ .
Two free masses on one spring: $\omega = \sqrt{k/\mu}$ with $\mu = \dfrac{m_1m_2}{m_1 + m_2}$ .
Two masses set moving: momentum is conserved. At the largest stretch both move with the centre-of-mass velocity, and the spring holds $\tfrac{1}{2}\mu v_{rel}^{2} = \tfrac{1}{2}kx_{max}^{2}$ .
Stacked blocks moving together: $T = 2\pi\sqrt{(M + m)/k}$ . Friction on the top block is $m \cdot \dfrac{kx}{M + m}$ , so it does not slip while the amplitude is at most $\mu g(M + m)/k$ .

Spring–mass period and reduced mass

T = 2\pi\sqrt{\frac{m}{k}} \qquad \omega = \sqrt{\frac{k}{\mu}},\ \ \mu = \frac{m_1m_2}{m_1 + m_2}

Worked example

A mass m on a spring oscillates with period 2 s. When 5 kg is added, the period becomes 3 s. Find m.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q117Moderate

Example 2 · Oscillations · Spring Systems and Restoring Forces

A mass

m

attached to free end of a spring executes SHM with a period of

1\text{ }s

. If the mass is increased by

3\text{ }kg

the period of oscillation increases by one second, the value of mass

m

kg

Two free masses use the reduced mass

With no wall, a spring between m₁ and m₂ oscillates with ω = √(k/μ), μ = m₁m₂/(m₁ + m₂). Using one mass or the total mass gives the wrong ω.

Gravity does not change a spring's period

Hanging the spring vertically shifts the equilibrium down by mg/k, but T = 2π√(m/k) is unchanged.

Equal amplitudes and equal top speeds give opposite ratios

For equal masses, v_max = A√(k/m). Equal amplitudes give v_max in the ratio √(k₁/k₂); equal v_max give amplitudes in the ratio √(k₂/k₁).

Drill 10 more on mass on a spring, and two masses on one spring

Concept 3 of 3: SHM from any restoring force or torque

A spring is only one way to get SHM. Displace any system a little, write the force (or torque) that pulls it back, and if it is a constant times the displacement, the motion is SHM. The constant plays the part of k, and ω² is that constant divided by the mass (or by the moment of inertia).

Definition

$F = -Cx$ gives $\omega = \sqrt{C/m}$ ; $\tau = -\kappa\theta$ gives $\omega = \sqrt{\kappa/I}$ .
Floating body of cross-section A in a liquid of density ρ: pushing it down by y adds buoyancy $\rho gAy$ , so $C = \rho gA$ and $T = 2\pi\sqrt{M/(\rho Ag)}$ . Since $M = \rho Ah$ (h = depth under the liquid), $T = 2\pi\sqrt{h/g}$ .
Tunnel through a uniform earth, along a diameter or any chord: the force along the tunnel is $-(mg/R)x$ , so $T = 2\pi\sqrt{R/g} \approx 84$ minutes.
From a potential $U(x)$ : C is $U''$ at the minimum. For $U = U_0(1 - \cos ax)$ , $C = U_0a^{2}$ .
Physical pendulum: $T = 2\pi\sqrt{I/(mgd)}$ , I about the pivot, d from the pivot to the centre of mass.
Dipole in a field E: $\tau = -pE\theta$ , so $T = 2\pi\sqrt{I/(pE)}$ .

Angular frequency from a restoring force or torque

F = -Cx \Rightarrow \omega = \sqrt{\frac{C}{m}} \qquad \tau = -\kappa\theta \Rightarrow \omega = \sqrt{\frac{\kappa}{I}}

Worked example

A wooden block of base area 50 cm² floats upright in water (density 1000 kg/m³) with 8 cm of it under water. It is pushed down a little and released. Find the period (g = 10 m/s²).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q3Moderate

Example 3 · Oscillations · Spring Systems and Restoring Forces

A cylindrical block of mass M and area of cross section A is floating in a liquid of density

\rho

and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is

\_\_\_\_

A physical pendulum uses I about the pivot

In T = 2π√(I/mgd), I is the moment of inertia about the pivot, found with the parallel-axis theorem. I about the centre of mass gives too short a period.

The tunnel period does not depend on the chord

The force along any chord through a uniform earth is −(mg/R)x, so every such tunnel gives T = 2π√(R/g), about 84 minutes.

A floating body's period depends on its depth under the liquid

T = 2π√(M/ρAg) equals 2π√(h/g), where h is the depth below the surface at rest. A denser liquid floats the block higher, so h is smaller and T is shorter.

Drill 7 more on shm from any restoring force or torque

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Cut springs and springs in series and in parallel
Cut and combined springs
$k \propto \frac{1}{l} \qquad \frac{1}{k_s} = \frac{1}{k_1} + \frac{1}{k_2} \qquad k_p = k_1 + k_2$
Mass on a spring, and two masses on one spring
Spring–mass period and reduced mass
$T = 2\pi\sqrt{\frac{m}{k}} \qquad \omega = \sqrt{\frac{k}{\mu}},\ \ \mu = \frac{m_1m_2}{m_1 + m_2}$
SHM from any restoring force or torque
Angular frequency from a restoring force or torque
$F = -Cx \Rightarrow \omega = \sqrt{\frac{C}{m}} \qquad \tau = -\kappa\theta \Rightarrow \omega = \sqrt{\frac{\kappa}{I}}$

Watch out for (9)

A block between two walls is parallel
→ Cut springs and springs in series and in parallel
Cutting a spring makes it stiffer
→ Cut springs and springs in series and in parallel
Gravity along an incline does not change ω
→ Cut springs and springs in series and in parallel
Two free masses use the reduced mass
→ Mass on a spring, and two masses on one spring
Gravity does not change a spring's period
→ Mass on a spring, and two masses on one spring
Equal amplitudes and equal top speeds give opposite ratios
→ Mass on a spring, and two masses on one spring
A physical pendulum uses I about the pivot
→ SHM from any restoring force or torque
The tunnel period does not depend on the chord
→ SHM from any restoring force or torque
A floating body's period depends on its depth under the liquid
→ SHM from any restoring force or torque

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