From acceleration to force
The kinematics of SHM gave us a single, sharp result: the acceleration of a particle executing simple harmonic motion is proportional to its displacement and oppositely directed, \(a(t)=-\omega^2 x(t)\). That statement is about geometry and time. The force law converts it into dynamics by asking the only question Newton ever asks — what force produces this acceleration?
Multiply the acceleration by the mass. By Newton's second law, the force on a particle of mass \(m\) executing SHM is
\(F(t)=ma=-m\omega^2 x(t)\).
The whole content of the force law for SHM is contained in this substitution. Nothing new has been assumed; the dynamics is simply the kinematics seen through \(F=ma\).
If \(a=-\omega^2 x\) is not yet automatic, revisit velocity and acceleration in SHM — the force law is built directly on that result.
The force law F = −kx
Group the constants \(m\) and \(\omega^2\) into a single quantity \(k\). NCERT §13.6 writes the force law in its standard form:
\(F(t)=-k\,x(t)\), where \(k=m\omega^2\).
The chain from kinematics to the named force constant is short and worth holding as one object rather than four separate equations.
| Step | Statement | Source |
|---|---|---|
| 1 · Kinematics | \(a=-\omega^2 x\) | SHM acceleration (NCERT Eq. 13.11) |
| 2 · Newton's 2nd law | \(F=ma=-m\omega^2 x\) | Apply \(F=ma\) |
| 3 · Define force constant | \(F=-kx\), with \(k=m\omega^2\) | NCERT Eq. 13.13, 13.14a |
| 4 · Solve for ω | \(\omega=\sqrt{k/m}\) | NCERT Eq. 13.14b |
| 5 · Period | \(T=\dfrac{2\pi}{\omega}=2\pi\sqrt{\dfrac{m}{k}}\) | From \(\omega=2\pi/T\) |
NCERT stresses that this is reversible. Going from the displacement equation \(x(t)=A\cos(\omega t+\phi)\) to the force law required two differentiations. Integrating the force law twice recovers the displacement equation. Displacement and force are two equivalent definitions of the same motion.
Restoring force geometry
The force law \(F=-kx\) is a straight line through the origin with negative slope. Plotting it makes the restoring property visible at a glance: wherever the particle is, the force arrow points back toward \(x=0\).
The second picture is the physical one: a mass on a frictionless surface, displaced from its mean position, with the restoring force always pulling it back. This is the prototype linear harmonic oscillator NCERT uses throughout the chapter.
The force constant k = mω²
The quantity \(k\) is the force constant of the oscillator — the force per unit displacement, measured in newtons per metre. From \(k=m\omega^2\) it carries two pieces of information at once: the inertia \(m\) of the oscillating body and the angular frequency \(\omega\) of the motion. A stiff system (large \(k\)) restores hard, oscillates fast, and has a short period.
| Quantity | Symbol & relation | SI unit | Note |
|---|---|---|---|
| Force constant | \(k=m\omega^2\) | N m\(^{-1}\) | Force per unit displacement; the slope magnitude of \(F\) vs \(x\) |
| Angular frequency | \(\omega=\sqrt{k/m}\) | rad s\(^{-1}\) | Depends only on \(k\) and \(m\) |
| Time period | \(T=2\pi\sqrt{m/k}\) | s | Independent of amplitude |
| Restoring force | \(F=-kx\) | N | Always toward the mean position |
ω and T from the force law
Rearranging \(k=m\omega^2\) gives the angular frequency directly, \(\omega=\sqrt{k/m}\), and since \(\omega=2\pi/T\) the period follows as \(T=2\pi\sqrt{m/k}\). These two expressions are the workhorses of nearly every numerical question in this chapter. The key qualitative reading is what they leave out.
- \(\omega\) depends only on \(k\) and \(m\). Heavier mass slows the oscillation (larger \(T\)); stiffer restoring force speeds it up (smaller \(T\)).
- \(\omega\) is independent of amplitude. A large swing and a small swing of the same oscillator share the same period. NCERT lists this in its Points to Ponder: the period of SHM does not depend on amplitude, energy, or phase constant.
- The force constant fixes the timescale. Two masses on springs of different \(k\) but identical \(m\) oscillate at frequencies in the ratio \(\sqrt{k_1/k_2}\).
The defining dynamical condition
NCERT makes a strong claim and it is examined directly: simple harmonic motion can be defined by its force law. A particle of mass \(m\) oscillating under a Hooke's-law restoring force \(F=-kx\) necessarily executes SHM with \(\omega=\sqrt{k/m}\). The Points to Ponder put it bluntly — every periodic motion is not simple harmonic; only periodic motion governed by \(F=-kx\) is simple harmonic.
So when a NEET problem hands you a system and asks "is the motion SHM?", the test is mechanical. Displace the body by \(x\), find the net force, and check whether it has the form \(F=-(\text{positive constant})\times x\). If it does, the motion is SHM and the constant is your \(k\). If the net force is proportional to \(x^2\), \(x^3\), \(\sin x\) (without the small-angle approximation), or anything other than the first power of \(x\), the motion is not SHM.
Linear vs non-linear restoring forces
The word linear is the whole game. SHM requires the restoring force to be proportional to the first power of displacement. A particle oscillating under \(F=-kx\) is called a linear harmonic oscillator. In the real world the force may contain small additional terms proportional to \(x^2\), \(x^3\), and so on; such systems are non-linear oscillators and their motion, though periodic, is not simple harmonic.
| Force law | Restoring? | Linear in x? | Is it SHM? |
|---|---|---|---|
| \(F=-kx\) | Yes | Yes | Yes — SHM |
| \(a=-10x\) | Yes | Yes | Yes — SHM (\(\omega^2=10\)) |
| \(a=-200x^2\) | Yes (for \(x>0\)) | No (\(x^2\)) | No — non-linear oscillator |
| \(a=+100x^3\) | No (drives outward) | No (\(x^3\)) | No |
| \(a=+0.7x\) | No (positive sign) | Yes | No — not restoring |
That table is essentially NCERT Exercise 13.6 in disguise, and it is a recurring NEET pattern: of several \(a\)-versus-\(x\) relations, exactly the ones of the form \(a=-(\text{positive})\,x\) are SHM.
Worked example — two-spring oscillator
Two identical springs of spring constant \(k\) are attached to a block of mass \(m\) and to fixed supports on either side. Show that when the mass is displaced from its equilibrium position, it executes simple harmonic motion, and find the period.
Displace the block a small distance \(x\) to the right. The left spring stretches by \(x\) and pulls the block left; the right spring compresses by \(x\) and pushes the block left. Both forces are restoring.
Each spring force: \(F_1=-kx\) (left spring) and \(F_2=-kx\) (right spring), both directed toward the mean position.
Net force: \(F=F_1+F_2=-2kx\). This has the form \(F=-(\text{positive constant})\times x\), so the motion is SHM with effective force constant \(k_{\text{eff}}=2k\).
Period: \(T=2\pi\sqrt{\dfrac{m}{k_{\text{eff}}}}=2\pi\sqrt{\dfrac{m}{2k}}\).
The method generalises to every "is it SHM?" problem: displace, sum the forces, read off the coefficient of \(x\). If the net force is \(-(\text{const})x\), the motion is simple harmonic and that constant is \(k_{\text{eff}}\).
Series and parallel combinations of springs, and the full \(T=2\pi\sqrt{m/k}\) toolkit, are covered in the spring-mass system. The energy stored in \(F=-kx\) is treated in energy in SHM.
The force law in one breath
- Start from \(a=-\omega^2 x\); apply \(F=ma\) to get \(F=-m\omega^2 x=-kx\).
- The force constant is \(k=m\omega^2\), measured in N m\(^{-1}\) — the force per unit displacement.
- Therefore \(\omega=\sqrt{k/m}\) and \(T=2\pi\sqrt{m/k}\).
- The negative sign makes \(F\) a restoring force — always directed toward the mean position.
- SHM is defined by \(F=-kx\): a linear restoring force. Forces in \(x^2\), \(x^3\), etc. give non-linear oscillators, not SHM.
- \(\omega\) and \(T\) depend only on \(k\) and \(m\), never on amplitude, energy, or phase.