Force law in SHM

Pull a trolley spring a little and let go: it snaps back, overshoots, comes back again, and keeps doing this — a clean, repeating dance that looks nothing like the messy way most pushed and pulled objects move. Why this particular back-and-forth, and not some other rhythm? The answer is not in how far you pulled it, or how hard. It is hidden in the rule the force obeys: at every instant, the spring pushes or pulls the block straight back toward the centre, and it pushes harder the farther the block has strayed. Write that rule as an equation and the whole rhythmic dance — the cosine wave, the fixed time period, the fact that a gently pulled spring and a hard-pulled spring tick at exactly the same rate — falls out as pure algebra. That rule is the force law in SHM: $F = -kx$ , Hooke's law dressed up as a law of motion. It looks almost too simple to carry so much weight, yet from this one line, combined with nothing more than Newton's second law, you can derive the entire kinematic picture of oscillations you met earlier in this chapter — the $-\omega^2 x$ acceleration, the angular frequency $\omega=\sqrt{k/m}$ , and the crown-jewel result for a mass on a spring, $T = 2\pi\sqrt{m/k}$ . This article builds that chain step by step: from the restoring-force law, through Newton's second law, to the time period of a loaded spring, with two fully worked numerical problems (a horizontal spring and a vertical one) in the CBSE numerical style, plus the traps examiners love to set around the minus sign, the independence from amplitude, and the difference between $k$ and $\omega$ .

Key takeaways

The defining rule of SHM is a force law, not just a graph shape. A particle is in SHM exactly when the net force on it is $F=-kx$ — proportional to displacement $x$ from the mean position, and always directed back toward it.
$k$ is the force constant (spring constant), SI unit N/m; it measures how "stiff" the restoring force is per unit displacement.
Newton's second law turns the force law into an acceleration law. $F=ma=-kx$ gives $a=-\dfrac{k}{m}x=-\omega^2 x$ , exactly the acceleration signature of SHM.
Matching coefficients hands you the angular frequency for free: $\omega=\sqrt{k/m}$ , and hence the time period $T=2\pi\sqrt{m/k}$ for any mass $m$ on a spring of force constant $k$ .
$T$ does not depend on amplitude $A$ — this isochronism is the single most tested conceptual fact in the topic; only $m$ and $k$ decide $T$ .
A vertical spring obeys the same law about a shifted mean position. Gravity only relocates where $x=0$ sits; the restoring-force constant $k$ , and therefore $T=2\pi\sqrt{m/k}$ , stay unchanged.
Key idea: whenever a net force obeys $F=-kx$ , Newton's second law forces the motion to be SHM with $\omega=\sqrt{k/m}$ and $T=2\pi\sqrt{m/k}$ , a period fixed by mass and stiffness alone, never by amplitude.

The core insight

A restoring force obeying $F=-kx$ combined with $F=ma$ forces $a=-\omega^2x$ with $\omega=\sqrt{k/m}$ , so any mass on a spring executes SHM with time period $T=2\pi\sqrt{m/k}$ , independent of amplitude.

Where it fits in the syllabus

This section sits inside the CBSE Class 11 "Oscillations" chapter, directly after periodic motion and the kinematic description of SHM ( $x=A\cos(\omega t+\phi)$ , $v=-A\omega\sin(\omega t+\phi)$ , $a=-\omega^2 x$ ) and just before energy in SHM and the simple pendulum. It is where the chapter switches from describing SHM kinematically to explaining it dynamically: given a restoring-force law $F=-kx$ and Newton's second law, why must the resulting motion be SHM at all, and what fixes its time period? The loaded-spring result $T=2\pi\sqrt{m/k}$ derived here is explicitly prescribed and is one of the most frequently examined numerical setups in the chapter; damped, forced and resonant oscillations of a spring are outside the current 2025-26 prescription and are not developed here.

The physical picture

Every SHM you have met — the spring, and later the simple pendulum — starts from the same physical demand: the net force must always try to restore the particle to one fixed point, the mean (equilibrium) position, and it must grow in direct proportion to how far the particle has been displaced from that point. This is the restoring-force law:

F = -kx \qquad (1)

where: $F$ is the net force on the particle (newton, N), $x$ is its displacement from the mean position (metre, m, positive on one side and negative on the other), and $k$ is the force constant (or spring constant), with SI unit N/m — a fixed property of the particular spring or restoring mechanism, not of the motion. The minus sign is the entire physical content of the word "restoring": when $x$ is positive (displaced to the right, say), $F$ is negative (pushes left, back toward zero); when $x$ is negative, $F$ is positive. The force is zero exactly at $x=0$ , the mean position, and it never points away from that position — a particle governed by $F=-kx$ can never run away to infinity; it is always being pulled home, harder the farther it strays.

Figure

A horizontal spring block system shown at three positions, compressed, at rest and stretched, with force arrows at each position always pointing back toward the centre and longer arrows at larger displacement, plus a small graph beside it of restoring force against displacement showing a straight line through the origin with negative slope

Now bring in dynamics. By Newton's second law, the net force on a particle of mass $m$ also equals $ma$ , so equation (1) becomes a statement about acceleration:

ma = -kx \quad\Longrightarrow\quad a = -\frac{k}{m}\,x \qquad (2)

where: $a$ is the instantaneous acceleration of the particle and $m$ is its mass (kg); the ratio $k/m$ is a fixed positive number set entirely by the physical system. Compare this with the acceleration signature of SHM you already derived kinematically, $a=-\omega^2 x$ . The two expressions have exactly the same form — acceleration proportional to $-x$ — so a particle obeying the force law $F=-kx$ is automatically in SHM, and reading off the coefficient gives the angular frequency directly:

\omega^2 = \frac{k}{m} \quad\Longrightarrow\quad \omega = \sqrt{\frac{k}{m}} \qquad (3)

where: $\omega$ is the angular frequency (rad/s), fixed once $k$ and $m$ are fixed — it does not depend on how the oscillation was started, how large the amplitude is, or the phase constant. This single relation is the bridge between the dynamics (a force law and a mass) and the kinematics (an angular frequency, and hence a time period) of the motion.

Deriving the time period of a loaded spring

Derivation AFrom the restoring-force law to the time period of a mass on a spring

The goal is to start from nothing but the restoring-force law of a spring, $F=-kx$ , and Newton's second law, $F=ma$ , and arrive at a single testable number: the time period $T$ of a mass $m$ oscillating on a spring of force constant $k$ . This is the central derivation of the topic because it shows, explicitly and without any extra physical assumption, why the motion must be periodic at all, and why that period cannot depend on how far the spring was initially stretched — a claim students often accept without seeing where it comes from.

Step 1. Write down the restoring-force law for the spring–mass system. A block of mass $m$ attached to a spring of force constant $k$ , displaced by $x$ from its natural (mean) length and released, experiences a spring force $F=-kx$ by Hooke's law. This is an experimental fact about the spring, valid as long as $x$ stays within the spring's elastic limit.

Step 2. Apply Newton's second law, $F=ma$ , to this same block. Since the only horizontal force acting is the spring force (the surface is taken as frictionless), $ma=-kx$ , which rearranges to $a=-\dfrac{k}{m}x$ . This is now a statement purely about how the block's acceleration depends on its own position — no reference to the spring force is needed once this equation is written down.

Step 3. Recognise the form of this equation. The kinematic definition of SHM, established earlier in the chapter from $x=A\cos(\omega t+\phi)$ , is precisely $a=-\omega^2x$ : acceleration directed opposite to displacement and proportional to it. Step 2's result, $a=-(k/m)x$ , has exactly this form, with the constant of proportionality $k/m$ playing the role of $\omega^2$ . Therefore the block's motion must be SHM, and by direct comparison, $\omega^2=k/m$ , so $\omega=\sqrt{k/m}$ .

Step 4. Convert the angular frequency into a time period using the standard relation between the two, $T=2\pi/\omega$ , which holds for any SHM regardless of what physical system produces it. Substituting $\omega=\sqrt{k/m}$ gives $T=\dfrac{2\pi}{\sqrt{k/m}}=2\pi\sqrt{\dfrac{m}{k}}$ .

Step 5. Check what this result does and does not depend on. Nowhere in steps 1 through 4 did the amplitude $A$ or the phase constant $\phi$ enter the calculation — $T=2\pi\sqrt{m/k}$ depends only on the mass $m$ and the force constant $k$ . This is the algebraic proof of isochronism: a spring pulled 1 cm and the same spring pulled 5 cm (both within the elastic limit) complete one oscillation in exactly the same time. A stiffer spring (larger $k$ ) gives a shorter $T$ ; a heavier block (larger $m$ ) gives a longer $T$ , matching the physical intuition that stiffness speeds up the oscillation and inertia slows it down.

Worked examples

Board level - angular frequency and time period of a horizontal spring-block system

A block of mass $m=0.5\ \text{kg}$ is attached to a horizontal spring of force constant $k=200\ \text{N/m}$ and set into oscillation on a frictionless surface. Find (a) the angular frequency, (b) the time period, and (c) the maximum acceleration of the block if the amplitude of oscillation is $A=0.02\ \text{m}$ .

Step 1. Angular frequency from $\omega=\sqrt{k/m}$ : $\omega=\sqrt{\dfrac{200}{0.5}}=\sqrt{400}=20\ \text{rad/s}$ .

Step 2. Time period from $T=2\pi/\omega$ : $T=\dfrac{2\pi}{20}=\dfrac{\pi}{10}\approx0.314\ \text{s}$ .

Step 3. Maximum acceleration occurs at maximum displacement, $x=A$ , since $a=-\omega^2x$ is largest in magnitude there: $a_{\max}=\omega^2A=400\times0.02=8\ \text{m/s}^2$ . Answer: $\omega=20\ \text{rad/s}$ , $T\approx0.314\ \text{s}$ , $a_{\max}=8\ \text{m/s}^2$ .

Competitive level - force constant, time period and restoring force of a vertical spring-mass system

A spring stretches by $0.05\ \text{m}$ when a mass of $2\ \text{kg}$ is hung from it and allowed to come to rest. The mass is then pulled down a further $0.03\ \text{m}$ from this new equilibrium and released. Using $g=9.8\ \text{m/s}^2$ , find (a) the force constant of the spring, (b) the time period of the resulting vertical oscillation, and (c) the magnitude of the restoring force acting on the mass at the instant of release.

Step 1. At the first equilibrium, the spring force balances the weight: $kx_0=mg$ , so $k=\dfrac{mg}{x_0}=\dfrac{2\times9.8}{0.05}=\dfrac{19.6}{0.05}=392\ \text{N/m}$ . Gravity has fixed the mean position of the subsequent oscillation, but it has not changed $k$ itself — the spring is still governed by $F=-kx$ about this new mean position.

Step 2. Time period: $T=2\pi\sqrt{\dfrac{m}{k}}=2\pi\sqrt{\dfrac{2}{392}}=2\pi\sqrt{0.0051}=2\pi\times0.0714\approx0.449\ \text{s}$ . This is identical in form to the horizontal case — a vertical spring obeys exactly the same $T=2\pi\sqrt{m/k}$ , since gravity only relocates $x=0$ , not $k$ or $m$ .

Step 3. Restoring force at release: the displacement from the new mean position at the moment of release is $x=0.03\ \text{m}$ (measured about the equilibrium found in Step 1, not about the spring's natural length), so $F=-kx=-392\times0.03=-11.76\ \text{N}$ , magnitude $11.76\ \text{N}$ , directed upward (back toward the equilibrium). Answer: $k=392\ \text{N/m}$ , $T\approx0.449\ \text{s}$ , restoring force $\approx11.76\ \text{N}$ upward at release. Timing note: under two minutes. The chain is $k=mg/x_0 \to T=2\pi\sqrt{m/k} \to F=-kx$ measured about the new mean position, not the spring's unstretched length — that last distinction is exactly where marks are lost under time pressure.

Common traps

Forgetting the negative sign in $F=-kx$ , or writing $F=kx$ — the sign is what makes the force restoring; drop it and the equation would describe a force pushing the particle further away, which cannot produce oscillation at all.
Believing $T=2\pi\sqrt{m/k}$ depends on gravity or on whether the spring is vertical or horizontal — it does not; gravity only shifts the mean (equilibrium) position for a vertical spring, while the restoring force about that new mean position is still $-kx$ with the same $k$ .
Assuming a larger amplitude gives a longer time period — in ideal SHM governed by $F=-kx$ , $T$ is fixed entirely by $m$ and $k$ ; amplitude never enters the formula.
Confusing the force constant $k$ (N/m, a property of the spring) with the angular frequency $\omega$ (rad/s, a property of the resulting motion) — they are related by $\omega=\sqrt{k/m}$ but are physically different quantities with different units.
In a vertical-spring problem, measuring displacement $x$ from the spring's unstretched (natural) length instead of from the new equilibrium position — this silently smuggles the weight $mg$ back into $F=-kx$ and gives a wrong restoring force.
Applying $F=-kx$ beyond the spring's elastic limit — the linear force law is Hooke's law, an idealization valid only for small enough displacements; stretched too far, a real spring no longer obeys $F=-kx$ and the resulting motion is no longer SHM.

Common trapThe single biggest misconception: mixing up the force constant k and the angular frequency omega

Students frequently write $\omega=k/m$ or plug $k$ directly into $T=2\pi/k$ , skipping the square root entirely. Remember the actual chain: the force law $F=-kx$ combined with $F=ma$ gives $a=-(k/m)x$ ; comparing this with the SHM signature $a=-\omega^2x$ shows that it is $\omega^2$ , not $\omega$ , that equals $k/m$ . Only after taking a square root do you get $\omega=\sqrt{k/m}$ , and only then does $T=2\pi/\omega=2\pi\sqrt{m/k}$ follow. Doubling $k$ does not double $\omega$ ; it multiplies $\omega$ by $\sqrt{2}$ , and it divides $T$ by $\sqrt{2}$ , not by $2$ . Always keep the square root explicit in every step of a numerical answer.

You should be able to…

State the restoring-force law $F=-kx$ and explain what makes it a restoring force, including the physical role of the minus sign.
Starting from $F=-kx$ and Newton's second law, derive $a=-\omega^2x$ and hence $\omega=\sqrt{k/m}$ .
Derive and apply $T=2\pi\sqrt{m/k}$ for a mass on a spring, both horizontal and vertical.
Explain, using the force law, why the time period of ideal SHM is independent of amplitude.
Correctly locate the mean (equilibrium) position for a vertical spring-mass system and measure displacement from there, not from the spring's unstretched length.
Solve for any one of $k$ , $m$ , $\omega$ , or $T$ given the other relevant quantities in a spring-mass numerical.

Ask the tutor: Want to walk through how $F=-kx$ turns step by step into $T=2\pi\sqrt{m/k}$ , or practice a few more spring-mass numericals with the AI tutor?

Why does a stiffer spring give a shorter time period instead of a longer one?

Think about what "stiffer" means physically: a larger force constant $k$ means the spring pushes back harder for the same displacement. A harder push means a larger acceleration back toward the mean position, so the block reverses direction and completes each swing faster. Algebraically, $\omega=\sqrt{k/m}$ grows with $k$ , and since $T=2\pi/\omega$ , a larger $\omega$ means a smaller $T$ . So yes — stiffer spring, faster oscillation, shorter time period. It is exactly the same logic as a tighter guitar string vibrating at a higher frequency.

If I pull the spring further before releasing it, why doesn't the block take longer to come back?

Because a larger displacement also means a larger restoring force at the start ( $F=-kx$ grows with $x$ ), which gives the block a larger initial acceleration and it ends up moving faster at every corresponding point of the swing. The extra distance to cover and the extra speed to cover it with cancel out exactly, so the time for one full oscillation, $T=2\pi\sqrt{m/k}$ , stays the same no matter the amplitude. This exact cancellation is a special feature of the linear restoring force $F=-kx$ ; it would not happen for a force law that was not directly proportional to $x$ .

Does the mass of the spring itself matter, or only the mass of the block?

In this chapter's scope, the spring is treated as ideal and massless, so only the block's mass $m$ appears in $T=2\pi\sqrt{m/k}$ . A real spring does have some mass, and a small correction (adding roughly a third of the spring's mass to $m$ ) is sometimes used in more advanced treatments, but that correction is beyond the CBSE Class 11 syllabus — for every problem here, use only the mass of the attached block.

Is this the same k as in the formula for elastic potential energy, half k x squared?

Yes, it is exactly the same force constant $k$ . The restoring force $F=-kx$ and the elastic potential energy $\tfrac12kx^2$ (met in the energy-in-SHM part of this chapter) both come from the identical spring, and $k$ carries the same value and the same SI unit, N/m, in both formulas. In fact, one can be obtained from the other: the potential energy is the work done against the restoring force in stretching the spring from $0$ to $x$ , which is why the same $k$ shows up in both places.