Pendulum and spring systems

Push a swing and let go, or pull a spring and release it — both settle into the same rhythmic back-and-forth called simple harmonic motion. The simple pendulum and the mass-on-a-spring are the two systems every Class 11 student must be able to analyse from scratch: write the restoring force, match it to $F=-ky$ , and read off the time period. A pendulum clock keeps steady time because $T=2\pi\sqrt{L/g}$ does not care how heavy the bob is; a spring balance reads the same whether it hangs still or bounces, because gravity only shifts where the mass sits, not how fast it oscillates about that point. This article builds both derivations from Newton's second law, cross-checks them with the energy picture, and works two full numerical examples — a swinging pendulum and a bouncing spring — so the formulas $T=2\pi\sqrt{L/g}$ and $T=2\pi\sqrt{m/k}$ stop being things to memorise and become things you can rebuild under exam pressure.

The core insight

Both the simple pendulum and the spring-mass system obey a restoring force proportional to displacement, $F=-ky_{\text{eff}}$ , which forces SHM with $T=2\pi\sqrt{L/g}$ for the pendulum and $T=2\pi\sqrt{m/k}$ for the spring, and neither period depends on the oscillating mass hanging on the pendulum string.

The core idea

Spring-mass system. An ideal spring obeying Hooke's law exerts a restoring force $F=-ky$ on a mass $m$ displaced by $y$ from equilibrium, where $k$ is the force constant (unit N/m). Applying Newton's second law:

m\frac{d^2y}{dt^2}=-ky \implies \frac{d^2y}{dt^2}+\omega^2y=0,\quad \omega^2=\frac{k}{m}

where: $m$ is the oscillating mass (kg), $k$ is the spring's force constant (N/m), $\omega$ is the angular frequency (rad/s). This is the standard SHM equation, so the motion repeats with time period

T=2\pi\sqrt{\frac{m}{k}}

This holds identically whether the spring is horizontal (gravity does no net work along the motion) or vertical (gravity only relocates the equilibrium point downward by $mg/k$ ; the restoring force measured from that new mean position is still $-ky'$ , so $T$ is unaffected).

Simple pendulum. A simple pendulum is an idealised point mass $m$ on a massless, inextensible string of length $L$ , swinging in a vertical plane. For a small angular displacement $\theta$ , the tangential restoring force is $-mg\sin\theta$ . Using the small-angle approximation $\sin\theta\approx\theta$ (valid in radians for roughly $\theta<10$ - $15°$ ):

F=-mg\theta=-mg\frac{y}{L},\qquad y=L\theta

This matches $F=-ky$ with effective constant $k=mg/L$ , so

m\frac{d^2y}{dt^2}=-\frac{mg}{L}y \implies \omega^2=\frac{g}{L},\qquad T=2\pi\sqrt{\frac{L}{g}}

where: $L$ is the string length (m), $g$ is local gravitational acceleration (m/s²). Crucially, $m$ has cancelled out entirely — a heavier bob does not swing slower. This is the same reasoning a pendulum clock relies on: as long as swings stay small-angle, $T$ depends only on $L$ and $g$ , never on amplitude or mass, so amplitude decay from air resistance barely changes the timekeeping.

Stiffness and inertia pull opposite ways. From $\omega^2=k/m$ , a stiffer spring (larger $k$ ) gives a shorter period since $T\propto1/\sqrt{k}$ , while a heavier mass (larger $m$ ) gives a longer period since $T\propto\sqrt{m}$ . Stiffness speeds the oscillation up; inertia slows it down — the two constants enter the formula with opposite exponents for a reason rooted in $F=ma$ itself.

Energy cross-check. Both systems store the same SHM energy budget, $E=\tfrac12m\omega^2A^2$ , split between kinetic energy at the mean position and potential energy at the extremes: elastic PE $\tfrac12ky^2$ for the spring, gravitational PE $mgL(1-\cos\theta)$ for the pendulum, which reduces for small $\theta$ to the same $\tfrac12ky^2$ form with $k=mg/L$ . Getting the same answer from force and from energy is a standard self-check before trusting a derived period.

Figure

Two panels comparing a spring mass system oscillating vertically about a shifted equilibrium and a simple pendulum swinging through a small angle theta about the vertical

Derivation ATime period of a simple pendulum from the small-angle restoring force

The goal is to show, starting only from Newton's second law and the small-angle approximation, that a simple pendulum of length $L$ executes SHM with time period $T=2\pi\sqrt{L/g}$ , a result completely independent of the bob's mass $m$ and of the swing amplitude, which is exactly why the formula is trusted for real pendulum clocks over many small swings.

Step 1. Set up the geometry. A point mass $m$ hangs from a massless string of length $L$ , displaced through angle $\theta$ from the vertical. Its arc-length displacement along the circular path is $y=L\theta$ , and gravity provides the only restoring force, acting tangent to the arc with magnitude $mg\sin\theta$ , directed back toward $\theta=0$ .

Step 2. Write Newton's second law along the arc. The tangential equation of motion is $m\dfrac{d^2y}{dt^2}=-mg\sin\theta$ , the minus sign showing the force always opposes the displacement.

Step 3. Apply the small-angle approximation. For $\theta$ measured in radians and kept below roughly $10$ - $15°$ , $\sin\theta\approx\theta$ . Substituting and using $\theta=y/L$ : $m\dfrac{d^2y}{dt^2}=-mg\dfrac{y}{L}$ , so the mass cancels from both sides, leaving $\dfrac{d^2y}{dt^2}=-\dfrac{g}{L}y$ .

Step 4. Match to the SHM equation $\ddot{y}+\omega^2y=0$ and read off $\omega^2=g/L$ . Since $T=2\pi/\omega$ , this gives $T=2\pi\sqrt{L/g}$ , depending only on $L$ and $g$ — never on $m$ or on amplitude, as long as the small-angle step remains valid. Verified: dimensional check, $[L/g]=\text{m}/(\text{m/s}^2)=\text{s}^2$ , so $\sqrt{L/g}$ has units of seconds, consistent with a time period.

Worked example

A simple pendulum of length $1.0\ \text{m}$ swings with small amplitude, and a spring-mass system undergoes vertical SHM. Take $g=9.8\ \text{m/s}^2$ .

Step 1. Pendulum time period: $T=2\pi\sqrt{\dfrac{L}{g}}=2\pi\sqrt{\dfrac{1.0}{9.8}}=2\pi\sqrt{0.1020}=2\pi\times0.3194=2.007\ \text{s}\approx2.0\ \text{s}$ , so the frequency is $f=1/T\approx0.5\ \text{Hz}$ .

Step 2. Length for a one-second period: from $T=2\pi\sqrt{L/g}$ , $L=\dfrac{gT^2}{4\pi^2}=\dfrac{9.8\times1^2}{39.48}=0.248\ \text{m}\approx24.8\ \text{cm}$ — this is the length a seconds-type pendulum design must hit for $T=1\ \text{s}$ one-way swings.

Step 3. Now the spring: a mass of $0.4\ \text{kg}$ hangs on a vertical spring and completes 20 oscillations in $16\ \text{s}$ , giving $T=16/20=0.8\ \text{s}$ . The force constant follows from $T=2\pi\sqrt{m/k}$ , rearranged as $k=\dfrac{4\pi^2m}{T^2}=\dfrac{4\times9.87\times0.4}{0.64}=\dfrac{15.79}{0.64}=24.68\ \text{N/m}$ .

Step 4. The static extension when the mass just hangs in equilibrium (spring force balances weight): $kx_0=mg \Rightarrow x_0=\dfrac{mg}{k}=\dfrac{0.4\times9.8}{24.68}=\dfrac{3.92}{24.68}=0.159\ \text{m}\approx15.9\ \text{cm}$ . This extension only relocates the mean position; the oscillation about it still obeys $T=2\pi\sqrt{m/k}$ , unchanged by gravity, exactly as the vertical-spring argument in the core idea predicts.

Final answers: pendulum $T\approx2.0\ \text{s}$ ( $L=0.248\ \text{m}$ for $T=1\ \text{s}$ ); spring $k\approx24.68\ \text{N/m}$ , static extension $\approx15.9\ \text{cm}$ .

Common traps

Using $T=2\pi\sqrt{L/g}$ for large swing angles — the small-angle approximation $\sin\theta\approx\theta$ breaks down beyond roughly $10$ - $15°$ , and the simple formula stops being accurate.
Believing a heavier bob makes a pendulum swing slower — mass cancels completely out of the pendulum equation; only $L$ and $g$ matter.
For a vertical spring, wrongly folding $g$ into the period formula (writing something like $T=2\pi\sqrt{(mg)/k}$ ) — gravity only shifts the equilibrium point; $T=2\pi\sqrt{m/k}$ is unchanged for either orientation.
Confusing the spring's force constant $k$ (N/m) with the propagation constant used later in wave equations — same letter, unrelated physical quantity.

Ask the tutor: Want a walkthrough of why the vertical-spring equilibrium shift leaves the period untouched, or a diagram-based derivation of the pendulum's small-angle step?

Why does the pendulum formula not include the mass of the bob at all?

Because mass appears on both sides of Newton's second law and cancels. The restoring force is $mg\sin\theta$ , but the inertia resisting acceleration is also $m$ , so in $m\ddot y=-mg\sin\theta$ the $m$ divides out before you even reach the small-angle step. Physically, a heavier bob feels a proportionally larger restoring force but also needs a proportionally larger force to accelerate it the same amount, so the two effects exactly cancel and only $L$ and $g$ survive in $T=2\pi\sqrt{L/g}$ .

If gravity does not appear in T = 2π√(m/k) for a vertical spring, why does the spring stretch differently once loaded?

Gravity does something, just not to the period. It sets a new mean (equilibrium) position, found from $kx_0=mg$ , and the mass oscillates symmetrically about that new point rather than about the spring's natural length. Once you measure displacement $y$ from this shifted mean, the net force is still exactly $-ky$ — the constant weight $mg$ and the constant extra spring tension $kx_0$ cancel at every instant, leaving pure SHM with $T=2\pi\sqrt{m/k}$ , identical to the horizontal case.

How small does the swing angle really need to be for the pendulum formula to work?

As a working rule, keep the amplitude under about $10°$ to $15°$ ; the error in $\sin\theta\approx\theta$ is under half a percent at $10°$ and grows quickly beyond that. For amplitudes larger than this, the true period is slightly longer than $2\pi\sqrt{L/g}$ and depends weakly on amplitude, but that correction uses a series expansion beyond the Class 11 syllabus — for exam purposes, always assume small-angle unless a large amplitude is explicitly stated as a conceptual exception.

Pendulum and spring systems

The core idea

Worked example

Common traps

Before this

Where this leads