Chapter 01 · Simple Harmonic Motion · Full book

Simple Harmonic Motion: The Heartbeat of Physics

Core intuition: Chapter Roadmap

A mass nudged from a stable equilibrium does not wander off—it returns, overshoots, and returns again. That single behaviour, repeated across springs, pendulums, atoms, and bridges, is simple harmonic motion. This chapter builds it from the ground up: first the motion, then its cause, then its energy, and finally the real-world complications of friction, forcing, and superposition.

Kinematics — the position function \(x = A\cos(\omega t + \varphi)\) and the velocity and acceleration that follow from it, including the amplitude, phase, and the relation \(v = \omega\sqrt{A^2 - x^2}\).
Dynamics: the restoring-force condition — why any linear restoring force \(F = -kx\) guarantees SHM with \(\omega = \sqrt{k/m}\), and why the period is independent of amplitude.
Energy in SHM — the constant exchange between kinetic and potential energy, the total \(E = \tfrac12 kA^2\), and the energy method as a shortcut to \(\omega\) for non-obvious oscillators.
Pendulums — the simple pendulum \(T = 2\pi\sqrt{L/g}\), the small-angle approximation and when it fails, and the physical pendulum with its equivalent simple length.
Spring–mass systems — springs in series and parallel, the vertical-spring equilibrium shift, and the two-body oscillator with its reduced mass \(\mu\).
Damped oscillations — the three regimes, the decay of amplitude and energy, and the quality factor \(Q\).
Forced oscillations and resonance — the steady-state response, the resonance peak, and the bandwidth that \(Q\) controls.
Superposition — combining SHMs along the same line and along perpendicular axes, and the beats that arise from nearby frequencies.
The projection connection — SHM as the shadow of uniform circular motion, a viewpoint that turns phase and timing problems into geometry.

Master the defining condition \(a = -\omega^2 x\), and every system in this chapter becomes the same problem wearing a different costume.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Mass-Independent Pendulum: A small bob of mass \(m\) hangs on a string of length \(L\) and swings as a pendulum. A second bob of mass \(100m\) hangs on an identical string. Both are released from the same small angle. They arrive back at the lowest point at exactly the same instant, every single swing, for as long as you watch. Why does mass—the very quantity that determines inertia and gravitational force—play no role in the period?
The Eternal Spring: A mass on a spring, in theory, oscillates forever once set in motion. In practice, every spring system you have ever seen eventually stops. And yet the mathematics of the ideal system is internally consistent, contradicts no law of physics, and gives predictions that match real experiments to extraordinary precision for long periods of time. What exactly is the ideal model ignoring, and how do we repair it?
The Resonance Catastrophe: Soldiers marching in step across a bridge can cause its destruction. The same bridge survives earthquakes, storms, and the weight of hundreds of people walking at random. A carefully designed bridge withstands a hundred times the force of the marching soldiers—yet it fails. How can a small, periodic force cause more damage than a large, random one?
The Tunnel Through the Earth: In Chapter we asked: if a frictionless tunnel were drilled straight through the centre of Earth, and a ball dropped in at one end, what would happen? The ball would oscillate back and forth, passing through the centre, never coming to rest. What force law produces this oscillation, and why does it have exactly the same period as a low-Earth orbit satellite?

Each question targets a trap that catches students every year.
By the end of this chapter, every one of them will be transparent.

Why SHM Deserves Its Own Chapter

Every chapter of this book has described a different category of motion. One-dimensional kinematics treated constant acceleration. Circular motion treated constant speed on a curve. Rotational mechanics extended Newton's laws to spinning bodies. Gravitation treated the long-range inverse-square force.

None of them prepared you for the motion of a pendulum clock, a tuning fork, the air column in a flute, the current in a circuit, the electric field of a light wave, or the vibrational modes of a chemical bond.

These are oscillations—motions that repeat themselves periodically, alternating direction, returning to their starting point, and repeating indefinitely. They are described by a completely different class of solutions to Newton's second law. The displacement is not linear in time, not quadratic, not a simple polynomial. It is sinusoidal.

The Unreasonable Ubiquity of SHM

What is surprising is not that oscillations exist, but that the same mathematical form describes them in such wildly different physical systems.

A mass on a spring, a pendulum for small angles, a floating cork bobbing on water, a torsion pendulum, an \(LC\) circuit, the vibration of atoms in a crystal, the normal modes of a drumhead—all of these systems, despite their obvious physical differences, obey exactly the same differential equation:

\[ \ddot{x} + \omega^2 x = 0 \]

This is the equation of simple harmonic motion (SHM). Its solutions are sines and cosines. Its parameter \(\omega\) is the angular frequency. Once you have seen why this equation governs a spring, you will immediately understand why it governs every other system in the list above.

Core intuition: The Reason SHM Is Everywhere

Near the minimum of any smooth potential energy function, the potential is approximately parabolic. A parabolic potential produces a linear restoring force. A linear restoring force produces SHM. This is not a coincidence specific to springs. It is a mathematical theorem: the Taylor expansion of any smooth \(U(x)\) about its minimum \(x_0\) begins \(U \approx U(x_0) + \tfrac{1}{2}U''(x_0)(x-x_0)^2\). The coefficient \(U''(x_0) \gt 0\) plays the role of a spring constant. Every small oscillation about a stable equilibrium is, to first order, SHM.

This is why a pendulum at small angles, despite obeying a completely different force law at large angles, reduces to SHM for small displacements. The gravitational restoring force near the bottom of its arc is approximately linear in the displacement.

Connections to Earlier Chapters

This chapter does not introduce an alien formalism. It extends tools you already possess.

Circular motion. The angular frequency \(\omega\) you used in circular motion (\(v = r\omega\), centripetal acceleration \(= r\omega^2\)) reappears here as the rate at which the phase of an oscillation advances. The connection is deeper than notation: we will show that SHM is the projection of uniform circular motion onto a diameter.

Energy methods. In Chapter you learned that the elastic potential energy of a spring is \(U = \tfrac{1}{2}kx^2\). The energy analysis of SHM is entirely a consequence of this expression combined with conservation of energy. You do not need the differential equation to derive the velocity–displacement relation; energy alone suffices.

Newton's laws. The equation \(\ddot{x} + \omega^2 x = 0\) is Newton's second law, \(F = ma\), applied to a system where the force is proportional to and opposite in direction from the displacement. Everything in this chapter is a consequence of one force law: \(F = -kx\).

Gravitation. The tunnel-through-Earth problem from the gravitation chapter leads directly to SHM, as we will show later in this chapter. The restoring force inside a uniform sphere is proportional to the displacement from the centre—precisely the condition for SHM.

What This Chapter Covers

We begin with the kinematics: the position, velocity, and acceleration functions that describe oscillatory motion. We then identify the physical condition—a linear restoring force— that produces this motion and derive the frequency formula. Energy in SHM follows, then a systematic treatment of physical systems: the simple pendulum, the physical pendulum, and spring combinations. We treat damping and resonance, superposition of oscillations, and the circular-motion connection. The chapter closes with the general result—SHM as the linearisation of any potential well—showing that every result here is a special case of a theorem about smooth energy functions.

Misconception alert

“SHM requires a spring.” Wrong. A spring is the canonical example, but SHM is a kinematic pattern, not a specific mechanism. Any system with a restoring force linear in displacement executes SHM. Pendulums (small angle), floating objects, torsion balances, and the gravitational tunnel all qualify. The spring is the simplest example, not the definition.
“The period of a pendulum depends on mass and amplitude.” Wrong (at small amplitudes). For small angles, the period \(T = 2\pi\sqrt{L/g}\) depends only on length and \(g\). Mass cancels because both inertia (\(m\)) and restoring force (\(mg\sin\theta \approx mg\theta\)) are proportional to \(m\). Amplitude independence holds only for small oscillations; for large amplitudes the period does depend on amplitude—a correction that JEE occasionally exploits.

✓ Checkpoint · Before We Begin

Name three physical systems, from completely different areas of physics, that all execute simple harmonic motion. For each, identify what plays the role of the “restoring force.”
(i) Mass on a spring: restoring force \(= -kx\), elastic force. (ii) Simple pendulum (small angle): restoring force \(\approx -mg\theta\), component of gravity. (iii) Object floating in a liquid displaced vertically: restoring force \(= -\rho A g\,y\), buoyancy. Many other valid answers exist.
A pendulum of length \(L\) has period \(T\) on Earth's surface. Without calculation, state how the period changes if: (a) the bob's mass is doubled, (b) the length is quadrupled, (c) the experiment is taken to a planet where \(g\) is halved.
(a) Period unchanged—mass cancels. (b) Period doubles—\(T \propto \sqrt{L}\), so \(\sqrt{4L} = 2\sqrt{L}\). (c) Period multiplies by \(\sqrt{2}\)—\(T \propto 1/\sqrt{g}\).
The differential equation \(\ddot{x} + \omega^2 x = 0\) has solutions \(x(t) = A\cos(\omega t + \phi)\). Without solving it, explain in words why the restoring force \(F = -kx\) should lead to a solution involving a cosine rather than a polynomial.
A polynomial \(x = at^2 + bt + c\) gives \(\ddot{x} = 2a = \text{const}\), which cannot equal \(-\omega^2 x\) for all \(t\). A cosine satisfies \(\ddot{x} = -\omega^2 \cos(\omega t)= -\omega^2 x\) identically—its second derivative is proportional to itself with a negative coefficient, which is exactly the algebraic form of the equation of motion.

Kinematics of SHM: Displacement, Velocity, and Acceleration

◈ The oscillation & its phasor

live bench

Drag amplitude, frequency and phase; x(t) redraws while the rotating phasor stays locked to it — SHM as the shadow of circular motion.

◈ loading the interactive bench…

↗ open in the Simulations page

Before asking why a system oscillates, we describe how it moves. The kinematic description of SHM begins with one experimental observation: when a mass on a spring is released from rest and its position is recorded as a function of time, the graph is a perfect cosine.

Not approximately a cosine. Not a cosine for small displacements only. Exactly a cosine, for the ideal spring at every amplitude.

This observation is our starting point. The dynamical justification—why this particular curve and not some other—comes in a later section.

The Position Function

Let \(x\) be the displacement of the oscillating mass from its equilibrium position, with positive \(x\) in one chosen direction. The most general position function consistent with periodic, cosine-shaped motion is:

\[ \boxed{x(t) = A\cos(\omega t + \phi)} \]

Three parameters completely specify the motion:

\(A \gt 0\) is the amplitude—the maximum displacement from equilibrium. The mass oscillates between \(x = -A\) and \(x = +A\).
\(\omega \gt 0\) is the angular frequency, measured in radians per second. It determines how rapidly the oscillation repeats.
\(\phi\) is the initial phase (or phase constant), determined by the initial conditions: where the mass is and how fast it is moving at \(t = 0\).

The motion is periodic with period: \[ T = \frac{2\pi}{\omega} \] and frequency: \[ f = \frac{1}{T} = \frac{\omega}{2\pi} \]

The factor of \(2\pi\) converts between the “rotational” counting of radians per second and the “ordinary” counting of complete cycles per second. \(\omega\) and \(f\) carry the same physical information in different units.

Initial Conditions and Phase

The phase constant \(\phi\) is not an independent physical property of the oscillator—it depends on when you start the clock.

If the mass starts at the positive extreme and is released from rest: \(x(0) = A\), \(v(0) = 0\). Substituting into Eq. : \(A\cos\phi = A\), so \(\phi = 0\), giving \(x(t) = A\cos(\omega t)\).

If the mass starts at equilibrium moving in the positive direction: \(x(0) = 0\), \(v(0) \gt 0\). Then \(A\cos\phi = 0\), so \(\phi = \pm\pi/2\). Choosing \(\phi = -\pi/2\) gives \(x(t) = A\cos(\omega t - \pi/2) = A\sin(\omega t)\), which is positive and increasing at \(t = 0\). ✓

The two special cases worth memorising: \[\begin{aligned} x(t) &= A\cos(\omega t) \qquad \text{[released from rest at } x = +A\text{]} \\[4pt] x(t) &= A\sin(\omega t) \qquad \text{[passes through equilibrium at } t=0 \text{, moving in }+x\text{]} \end{aligned}\]

Velocity

Differentiate Eq. once:

\[ v(t) = \dot{x} = -A\omega\sin(\omega t + \phi) \]

Key observations:

The maximum speed is \(v_{\max} = A\omega\), achieved when \(\sin(\omega t + \phi) = \pm 1\), i.e. when \(x = 0\) (equilibrium).
The velocity is zero when \(x = \pm A\) (the turning points).
Velocity leads displacement by a quarter period: when displacement is at its peak, velocity is zero; when displacement passes through zero (going positive), velocity is at its negative maximum.

Acceleration

Differentiate once more:

\[ a(t) = \ddot{x} = -A\omega^2\cos(\omega t + \phi) = -\omega^2 x(t) \]

The last equality is crucial. The acceleration is proportional to displacement and opposite in sign. This is not a derived result in kinematics—it is the defining algebraic property of SHM. When you see \(a \propto -x\), you are looking at simple harmonic motion.

The maximum acceleration is \(a_{\max} = A\omega^2\), occurring at the turning points \(x = \pm A\) (where the restoring force is largest). At equilibrium, the acceleration is zero.

Core intuition: Phase Relationships in SHM

In SHM with \(x = A\cos(\omega t + \phi)\):

Velocity leads displacement by \(\pi/2\) (a quarter cycle): speed is maximum when displacement is zero.
Acceleration leads velocity by \(\pi/2\), i.e.\ acceleration is anti-phase with displacement by \(\pi\): acceleration is most negative when displacement is most positive.

These relationships are fixed by the form of the solution—they do not depend on initial conditions.

Velocity as a Function of Position

For energy problems and examination questions it is often more useful to have \(v\) as a function of \(x\) rather than \(t\). Using \(\sin^2\theta + \cos^2\theta = 1\):

\[ \left(\frac{x}{A}\right)^2 + \left(\frac{v}{A\omega}\right)^2 = 1 \]

\[ \boxed{v = \pm\,\omega\sqrt{A^2 - x^2}} \]

The \(\pm\) reflects the fact that at a given position \(x\) (inside the amplitude), the mass may be moving in either direction. At \(x = 0\): \(v = \pm A\omega\) (maximum speed). At \(x = \pm A\): \(v = 0\) (turning points).

This result is also derivable directly from energy conservation, as we will see in a later section.

Summary of Kinematic Quantities

Fig 9.1 Displacement, velocity and acceleration in SHM. Velocity leads displacement by a quarter period; acceleration is exactly out of phase, a = −ω²x.

Quantity	Formula	Maximum value
Displacement	\(x = A\cos(\omega t + \phi)\)	\(A\) at turning points
Velocity	\(v = -A\omega\sin(\omega t + \phi)\)	\(A\omega\) at \(x=0\)
Acceleration	\(a = -A\omega^2\cos(\omega t+\phi)\)	\(A\omega^2\) at \(x=\pm A\)

Example 1

A particle executes SHM with amplitude \(A = 0.10\,\mathrm{m}\) and period \(T = 2.0\,\mathrm{s}\). At \(t = 0\) the particle is at \(x = 0.05\,\mathrm{m}\) moving in the positive direction. (a) Find \(\omega\), \(\phi\), and write down \(x(t)\). (b) Find the speed and acceleration at \(t = 0\). (c) Find the first time after \(t = 0\) when the speed is maximum.

Solution. (a) \(\omega = 2\pi/T = 2\pi/2.0 = \pi\,\mathrm{rad\,s^{-1}}\). At \(t = 0\): \(x(0) = A\cos\phi = 0.05\,\mathrm{m}\), so \(\cos\phi = 0.05/0.10 = 1/2\), giving \(\phi = \pm\pi/3\). The velocity at \(t = 0\) must be positive: \(v(0) = -A\omega\sin\phi \gt 0 \Rightarrow \sin\phi \lt 0\). Since \(\cos\phi = 1/2 \gt 0\) and \(\sin\phi \lt 0\), the phase lies in the fourth quadrant: \(\phi = -\pi/3\). \[ x(t) = 0.10\cos\!\left(\pi t - \frac{\pi}{3}\right)\,\mathrm{m} \] (b) Speed at \(t = 0\): \[ v(0) = -A\omega\sin\phi = -(0.10)(\pi)\sin(-\pi/3) = (0.10)(\pi)\frac{\sqrt{3}}{2} \approx 0.272\,\mathrm{m\,s^{-1}} \] Alternatively, Eq. : \(v = \omega\sqrt{A^2 - x^2} = \pi\sqrt{0.01 - 0.0025} = \pi\sqrt{0.0075} \approx 0.272\,\mathrm{m\,s^{-1}}\). Acceleration at \(t = 0\): \[ a(0) = -\omega^2 x(0) = -\pi^2 \times 0.05 \approx -0.493\,\mathrm{m\,s^{-2}} \] The negative sign indicates the acceleration points in the \(-x\) direction, as expected: the particle is displaced in \(+x\) so the restoring force pulls it back. (c) Maximum speed occurs at \(x = 0\), i.e. when \(\cos(\pi t - \pi/3) = 0\): \[ \pi t - \frac{\pi}{3} = \frac{\pi}{2} + n\pi, \quad n = 0, 1, 2, \ldots \] The first solution (after \(t = 0\), with \(n = 0\)): \[ t = \frac{1}{\pi}\left(\frac{\pi}{2} + \frac{\pi}{3}\right) = \frac{1}{\pi} \cdot \frac{5\pi}{6} = \frac{5}{6}\,\mathrm{s} \]

✓ Checkpoint · Kinematics of SHM

A particle in SHM has amplitude \(A = 0.20\,\mathrm{m}\) and \(\omega = 4\,\mathrm{rad\,s^{-1}}\). Find: (a) the maximum speed, (b) the maximum acceleration, (c) the speed when \(x = 0.12\,\mathrm{m}\).
(a) \(v_{\max} = A\omega = 0.80\,\mathrm{m\,s^{-1}}\). (b) \(a_{\max} = A\omega^2 = 3.2\,\mathrm{m\,s^{-2}}\). (c) \(v = \omega\sqrt{A^2 - x^2} = 4\sqrt{0.04 - 0.0144} = 4\sqrt{0.0256} = 4 \times 0.16 = 0.64\,\mathrm{m\,s^{-1}}\).
A particle executes SHM with \(x(t) = 3\sin(2\pi t)\,\mathrm{cm}\). At what positions is the speed equal to half its maximum value?
Maximum speed \(= A\omega = 3 \times 2\pi\,\mathrm{cm\,s^{-1}}\). Half this speed: \(\omega\sqrt{A^2 - x^2} = A\omega/2 \Rightarrow A^2 - x^2 = A^2/4 \Rightarrow x = \pm A\sqrt{3}/2 = \pm 3\sqrt{3}/2 \approx \pm 2.6\,\mathrm{cm}\).
The position of a particle is \(x = 5\cos(3t + \pi/4)\,\mathrm{cm}\). Find the acceleration at \(t = 0\) and state its direction.
\(a(0) = -\omega^2 x(0) = -9 \times 5\cos(\pi/4) = -9 \times 5/\sqrt{2} \approx -31.8\,\mathrm{cm\,s^{-2}}\). Direction: negative \(x\)-axis (toward equilibrium, opposing the positive initial displacement).

Dynamics of SHM: The Restoring Force Condition

The kinematic description of the previous section told us what the motion looks like. Now we ask: what force law produces it?

The answer is short and precise.

From Kinematics to Force

Fig 9.2 The defining condition of SHM: a restoring force proportional to displacement, F = −kx — a straight line of slope −k.

We showed in Section that the defining algebraic property of SHM is: \[ a = -\omega^2 x \]

Multiplying both sides by the mass \(m\): \[ F = ma = -m\omega^2 x \]

Writing \(k \equiv m\omega^2\):

\[ \boxed{F = -kx} \]

A system executes SHM if and only if the net restoring force on it is proportional to and opposite in direction from its displacement from equilibrium. The constant of proportionality \(k\) has SI units of \(\mathrm{N\,m^{-1}}\).

The angular frequency is then:

\[ \boxed{\omega = \sqrt{\frac{k}{m}}} \]

and the period:

\[ \boxed{T = 2\pi\sqrt{\frac{m}{k}}} \]

Core intuition: The Condition for SHM

A system executes simple harmonic motion if and only if the restoring force is linear in the displacement: \[ F_{\text{restoring}} = -kx \] This single condition implies sinusoidal motion, a well-defined angular frequency \(\omega = \sqrt{k/m}\), and a period independent of amplitude.

Derivation from Newton's Second Law

We can also run the argument forward: start from \(F = -kx\) and derive the motion.

Birth of the formula: SHM from \(F = -kx\)

Apply Newton's second law to a mass \(m\) subject to \(F = -kx\): \[ m\ddot{x} = -kx \quad\Longrightarrow\quad \ddot{x} + \frac{k}{m}x = 0 \] Define \(\omega^2 \equiv k/m\): \[ \ddot{x} + \omega^2 x = 0 \] This is a second-order linear ordinary differential equation with constant coefficients. Its general solution is: \[ x(t) = C_1\cos(\omega t) + C_2\sin(\omega t) \] which can always be written as \(A\cos(\omega t + \phi)\) by choosing \[ A = \sqrt{C_1^2 + C_2^2}, \qquad \tan\phi = -\frac{C_2}{C_1} \] The constants \(C_1\) and \(C_2\) (or equivalently \(A\) and \(\phi\)) are fixed by the initial conditions \(x(0)\) and \(v(0)\). The key result: the frequency \(\omega\) is determined by \(k/m\) alone, not by the initial conditions. The amplitude and phase depend on how the oscillation was started; the frequency does not.

Why Period Is Independent of Amplitude

This is a profound property unique to the linear restoring force. If you double the amplitude, you double the displacement at every instant—but this doubles the force (since \(F = -kx\)), which doubles the acceleration. A larger acceleration compensates exactly for the larger distance to be covered, leaving the period unchanged.

For any nonlinear restoring force, this compensation is not exact and the period depends on amplitude. A simple pendulum at large angles is a classic example: the period increases slightly with amplitude because \(\sin\theta \lt \theta\) makes the restoring force weaker than linear, so the pendulum takes longer.

Applications: Identifying the Frequency

In practice, you identify SHM by finding the restoring force and reading off the coefficient: \[ F = -(\text{effective }k)\cdot x \quad\Longrightarrow\quad \omega = \sqrt{\frac{\text{effective }k}{m}} \]

Three standard cases that appear repeatedly in JEE and NEET:

Horizontal spring. A mass \(m\) on a frictionless surface attached to a spring of stiffness \(k\): \[ F = -kx \quad\Longrightarrow\quad \omega = \sqrt{k/m} \] No subtlety here; gravity plays no role in the horizontal direction.

Vertical spring (equilibrium first). A mass \(m\) hangs from a spring of natural length \(L_0\) and stiffness \(k\). At equilibrium the spring stretches by \(\delta = mg/k\). Measure displacement \(y\) from this equilibrium position. The net force is: \[ F = -k(L_0 + \delta + y) + mg = -k\delta - ky + mg = -ky \] (since \(k\delta = mg\)). The result: measuring from the equilibrium position, gravity cancels entirely, and the frequency is still \(\omega = \sqrt{k/m}\). Gravity only shifts the equilibrium position; it does not change the frequency.

The tunnel through Earth. This resolves the fourth perplexing question from the chapter opener. A uniform sphere of mass \(M_E\) and radius \(R_E\) exerts a gravitational force on an internal mass \(m\) at depth \(r \lt R_E\) of: \[ F = -\frac{GM_E m}{R_E^3}\,r \] This is linear in \(r\) and directed toward the centre: a restoring force with effective spring constant \(k_{\text{eff}} = GM_Em/R_E^3\). The period: \[ T = 2\pi\sqrt{\frac{m}{k_{\text{eff}}}} = 2\pi\sqrt{\frac{R_E^3}{GM_E}} \] Substituting \(GM_E/R_E^2 = g\): \[ T = 2\pi\sqrt{\frac{R_E}{g}} \approx 2\pi\sqrt{\frac{6.4\times10^6}{9.8}} \approx 5060\,\mathrm{s} \approx 84\,\mathrm{min} \] This is exactly the period of a low-altitude circular orbit. The coincidence is not accidental: both the orbiting satellite and the tunneling mass sample the full mass of Earth in the same way, and both periods depend on the same ratio \(R_E/g\).

Example 2

A block of mass \(m = 0.50\,\mathrm{kg}\) is attached to a vertical spring of spring constant \(k = 200\,\mathrm{N\,m^{-1}}\). The block is pulled \(4.0\,\mathrm{cm}\) below the equilibrium position and released from rest. (a) Find the angular frequency, period, and amplitude. (b) Find the maximum speed and maximum acceleration. (c) Find the net force on the block when it is \(2.0\,\mathrm{cm}\) above the equilibrium position.

Solution. (a) \(\omega = \sqrt{k/m} = \sqrt{200/0.50} = \sqrt{400} = 20\,\mathrm{rad\,s^{-1}}\). \(T = 2\pi/\omega = 2\pi/20 = \pi/10 \approx 0.314\,\mathrm{s}\). The amplitude equals the initial displacement from equilibrium: \(A = 4.0\,\mathrm{cm} = 0.040\,\mathrm{m}\) (the block is released from rest, so it starts at a turning point). (b) \(v_{\max} = A\omega = 0.040 \times 20 = 0.80\,\mathrm{m\,s^{-1}}\). \(a_{\max} = A\omega^2 = 0.040 \times 400 = 16\,\mathrm{m\,s^{-2}}\). (c) At \(x = +2.0\,\mathrm{cm}\) above equilibrium, the net restoring force is: \[ F = -kx = -200 \times 0.020 = -4.0\,\mathrm{N} \] The force is \(4.0\,\mathrm{N}\) directed downward (toward equilibrium), regardless of gravity. Gravity is already accounted for in defining the equilibrium position; measuring \(x\) from equilibrium absorbs the gravitational term exactly.

✓ Checkpoint · Dynamics of SHM

A particle of mass \(m\) is subject to a restoring force \(F = -8x\,\mathrm{N}\) (with \(x\) in metres). Find the period of oscillation if \(m = 2\,\mathrm{kg}\).
\(\omega = \sqrt{k/m} = \sqrt{8/2} = 2\,\mathrm{rad\,s^{-1}}\). \(T = 2\pi/\omega = \pi \approx 3.14\,\mathrm{s}\).
A mass \(m\) hangs from a spring of stiffness \(k\). A student argues: “In the vertical spring problem gravity pulls the mass down so it must affect the frequency.” Explain clearly why this is wrong.
When displacement is measured from the equilibrium position, the gravitational force \(mg\) is exactly cancelled by the extra spring force \(k\delta = mg\) at equilibrium. The net force is purely \(-ky\), independent of gravity. Gravity determines where equilibrium is, not how fast the system oscillates about it.
Two springs of stiffness \(k_1\) and \(k_2\) are connected in series and support a mass \(m\) vertically. Without deriving the formula, identify what effective spring constant \(k_{\text{eff}}\) you would expect and write down the period. (Hint: for a given applied force, how much does each spring stretch?)
For springs in series, the same force \(F\) stretches each spring: \(x_1 = F/k_1\), \(x_2 = F/k_2\), total \(x = F(1/k_1 + 1/k_2)\). So \(1/k_{\text{eff}} = 1/k_1 + 1/k_2\), i.e.\ \(k_{\text{eff}} = k_1k_2/(k_1+k_2)\). Period: \(T = 2\pi\sqrt{m(k_1+k_2)/(k_1k_2)}\).

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Frequency of a spring–mass system

A block of mass \(m = 0.50\,\mathrm{kg}\) rests on a frictionless surface, attached to a spring of stiffness \(k = 200\,\mathrm{N/m}\). Find the angular frequency, the period, and the frequency of small oscillations.

Solution. Given: \(m = 0.50\,\mathrm{kg}\), \(k = 200\,\mathrm{N/m}\).
Find: \(\omega\), \(T\), \(f\).
Setup: For a horizontal spring–mass system the restoring force is \(F = -kx\), so \(\omega = \sqrt{k/m}\) by .
Solve: \[\begin{aligned} \omega &= \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.50}} = \sqrt{400} = 20\,\mathrm{rad/s},\\ T &= \frac{2\pi}{\omega} = \frac{2\pi}{20} = 0.314\,\mathrm{s},\\ f &= \frac{1}{T} = \frac{\omega}{2\pi} = \frac{20}{2\pi} = 3.18\,\mathrm{Hz}. \end{aligned}\] Answer: \(\boxed{\omega = 20\,\mathrm{rad/s},\; T = 0.314\,\mathrm{s},\; f = 3.18\,\mathrm{Hz}}\)
Check: Dimensionally \([k/m] = (\mathrm{N/m})/\mathrm{kg} = \mathrm{s^{-2}}\), so \(\sqrt{k/m}\) has units of \(\mathrm{s^{-1}}\) \(\checkmark\). And \(fT = 3.18 \times 0.314 = 1.00\) as required. \(\checkmark\)

Example 2 · Reading off amplitude, speed, and acceleration

A particle oscillates according to \[ x(t) = 0.08\cos(50t + \pi/3)\,\mathrm{m} \qquad (t \text{ in seconds}). \] State the amplitude, the maximum speed, the maximum acceleration, and the period.

Solution. Given: \(x(t) = 0.08\cos(50t + \pi/3)\) (metres, seconds).
Find: \(A\), \(v_{\max}\), \(a_{\max}\), \(T\).
Setup: Compare with \(x = A\cos(\omega t + \varphi)\): the coefficient is \(A\), the factor multiplying \(t\) is \(\omega\). Then \(v_{\max} = \omega A\) and \(a_{\max} = \omega^2 A\) from and .
Solve: \[\begin{aligned} A &= 0.08\,\mathrm{m}, \qquad \omega = 50\,\mathrm{rad/s},\\ v_{\max} &= \omega A = 50 \times 0.08 = 4.0\,\mathrm{m/s},\\ a_{\max} &= \omega^2 A = 50^2 \times 0.08 = 200\,\mathrm{m/s^2},\\ T &= \frac{2\pi}{\omega} = \frac{2\pi}{50} = 0.126\,\mathrm{s}. \end{aligned}\] Answer: \(\boxed{A = 0.08\,\mathrm{m},\; v_{\max} = 4.0\,\mathrm{m/s},\; a_{\max} = 200\,\mathrm{m/s^2},\; T = 0.126\,\mathrm{s}}\)
Check: The maxima are linked by \(a_{\max} = \omega\,v_{\max} = 50 \times 4.0 = 200\,\mathrm{m/s^2}\) \(\checkmark\), an internal consistency that does not use \(A\) again. \(\checkmark\)

Example 3 · Angular frequency from \(a = -\omega^2 x\)

At one instant a particle in SHM is at \(x = 0.02\,\mathrm{m}\) and has acceleration \(a = -8\,\mathrm{m/s^2}\) (directed toward equilibrium). The amplitude is \(A = 0.05\,\mathrm{m}\). Find \(\omega\), the period, and the maximum speed.

Solution. Given: \(x = 0.02\,\mathrm{m}\), \(a = -8\,\mathrm{m/s^2}\), \(A = 0.05\,\mathrm{m}\).
Find: \(\omega\), \(T\), \(v_{\max}\).
Setup: The defining relation of SHM is \(a = -\omega^2 x\) (); solve for \(\omega\), then use the standard period and \(v_{\max} = \omega A\).
Solve: \[\begin{aligned} \omega^2 &= -\frac{a}{x} = -\frac{-8}{0.02} = 400\,\mathrm{s^{-2}} \;\Rightarrow\; \omega = 20\,\mathrm{rad/s},\\ T &= \frac{2\pi}{\omega} = 0.314\,\mathrm{s},\\ v_{\max} &= \omega A = 20 \times 0.05 = 1.0\,\mathrm{m/s}. \end{aligned}\] Answer: \(\boxed{\omega = 20\,\mathrm{rad/s},\; T = 0.314\,\mathrm{s},\; v_{\max} = 1.0\,\mathrm{m/s}}\)
Check: At the extreme \(x = A = 0.05\,\mathrm{m}\) the acceleration would be \(\omega^2 A = 400 \times 0.05 = 20\,\mathrm{m/s^2}\), larger in magnitude than the \(8\,\mathrm{m/s^2}\) found at the smaller displacement \(0.02\,\mathrm{m}\), as it must be since \(|a| \propto |x|\). \(\checkmark\)

Example 4 · The simple pendulum and the seconds pendulum

A simple pendulum has length \(L = 1.0\,\mathrm{m}\). Using \(g = 9.8\,\mathrm{m/s^2}\), find its period. What length gives a “seconds pendulum,” for which \(T = 2.0\,\mathrm{s}\)?

Solution. Given: \(L = 1.0\,\mathrm{m}\), \(g = 9.8\,\mathrm{m/s^2}\); target \(T = 2.0\,\mathrm{s}\).
Find: period of the \(1\,\mathrm{m}\) pendulum; length for \(T = 2.0\,\mathrm{s}\).
Setup: Small-amplitude period \(T = 2\pi\sqrt{L/g}\) (); invert for \(L\).
Solve: \[\begin{aligned} T &= 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1.0}{9.8}} = 2.01\,\mathrm{s},\\ L_{\text{sec}} &= \frac{g T^2}{4\pi^2} = \frac{9.8 \times (2.0)^2}{4\pi^2} = \frac{39.2}{39.48} = 0.993\,\mathrm{m}. \end{aligned}\] Answer: \(\boxed{T = 2.01\,\mathrm{s},\quad L_{\text{sec}} = 0.993\,\mathrm{m}}\)
Check: The \(1\,\mathrm{m}\) pendulum already has \(T \approx 2.0\,\mathrm{s}\), so the seconds-pendulum length should sit just below \(1\,\mathrm{m}\)—and \(0.993\,\mathrm{m}\) does. \(\checkmark\)

Example 5 · Energy and maximum speed

A \(0.20\,\mathrm{kg}\) mass on a spring of constant \(k = 80\,\mathrm{N/m}\) oscillates with amplitude \(A = 0.10\,\mathrm{m}\). Find the total mechanical energy and the maximum speed.

Solution. Given: \(m = 0.20\,\mathrm{kg}\), \(k = 80\,\mathrm{N/m}\), \(A = 0.10\,\mathrm{m}\).
Find: \(E\), \(v_{\max}\).
Setup: Total energy \(E = \tfrac12 k A^2\) (); at equilibrium all of it is kinetic, so \(\tfrac12 m v_{\max}^2 = E\).
Solve: \[\begin{aligned} E &= \tfrac12 k A^2 = \tfrac12 \times 80 \times (0.10)^2 = 0.40\,\mathrm{J},\\ v_{\max} &= \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 0.40}{0.20}} = 2.0\,\mathrm{m/s}. \end{aligned}\] Answer: \(\boxed{E = 0.40\,\mathrm{J},\; v_{\max} = 2.0\,\mathrm{m/s}}\)
Check: Independently, \(\omega = \sqrt{k/m} = \sqrt{80/0.20} = 20\,\mathrm{rad/s}\), so \(v_{\max} = \omega A = 20 \times 0.10 = 2.0\,\mathrm{m/s}\)—the kinematic route agrees with the energy route. \(\checkmark\)

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Reading the Motion
A particle's position is given by \(x(t) = 6\cos(3\pi t + \pi/6)\) cm. Find: (a) amplitude, (b) angular frequency, (c) period, (d) initial position, (e) initial velocity.
(a) \(A = 6\,\mathrm{cm}\). (b) \(\omega = 3\pi\,\mathrm{rad\,s^{-1}}\). (c) \(T = 2\pi/\omega = 2/3\,\mathrm{s}\). (d) \(x(0) = 6\cos(\pi/6) = 3\sqrt{3} \approx 5.20\,\mathrm{cm}\). (e) \(v(0) = -A\omega\sin(\pi/6) = -6 \times 3\pi \times \tfrac12 = -9\pi \approx -28.3\,\mathrm{cm\,s^{-1}}\).
Frequency from Force
A particle of mass \(0.40\,\mathrm{kg}\) experiences a restoring force \(F = -16x\,\mathrm{N}\) (with \(x\) in metres). Find \(\omega\), \(f\), and \(T\).
\(k_\text{eff} = 16\,\mathrm{N\,m^{-1}}\). \(\omega = \sqrt{k/m} = \sqrt{16/0.40} = \sqrt{40} \approx 6.32\,\mathrm{rad\,s^{-1}}\). \(f = \omega/2\pi \approx 1.01\,\mathrm{Hz}\). \(T = 1/f \approx 0.994\,\mathrm{s}\).
Energy at a Given Position
A spring–mass system (\(k = 80\,\mathrm{N\,m^{-1}}\), \(m = 0.50\,\mathrm{kg}\)) has amplitude \(A = 10\,\mathrm{cm}\). Find the kinetic energy when \(x = 6\,\mathrm{cm}\).
\(E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(80)(0.10)^2 = 0.40\,\mathrm{J}\). \(U = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(80)(0.06)^2 = 0.144\,\mathrm{J}\). \(K = E - U = 0.256\,\mathrm{J}\).
Physical Pendulum — Off-Centre Rod
A thin uniform rod of mass \(M = 0.60\,\mathrm{kg}\) and length \(L = 1.2\,\mathrm{m}\) is pivoted about a horizontal axis through a point one-quarter of its length from one end. Find the period of small oscillations. (\(g = 9.8\,\mathrm{m\,s^{-2}}\).)
Pivot at \(L/4\) from one end; CM at \(L/2\), so \(d = L/4 = 0.30\,\mathrm{m}\). \(I = I_\text{cm} + Md^2 = ML^2/12 + M(L/4)^2 = ML^2(1/12 + 1/16) = \tfrac{7}{48}ML^2\). \(T = 2\pi\sqrt{I/(Mgd)} = 2\pi\sqrt{\tfrac{7}{48}ML^2/(Mg L/4)} = 2\pi\sqrt{7L/(12g)} = 2\pi\sqrt{7(1.2)/(12 \times 9.8)} \approx 1.68\,\mathrm{s}\).
Superposition of Three SHMs
Three collinear SHMs combine: \(x_1 = 2\cos\omega t\), \(x_2 = 3\cos(\omega t + \pi/3)\), \(x_3 = 4\cos(\omega t + 2\pi/3)\) (cm). Find the amplitude and phase of the resultant.
Phasor components: \(X = 2 + 3\cos60^\circ + 4\cos120^\circ = 2 + 1.5 - 2 = 1.5\,\mathrm{cm}\); \(Y = 3\sin60^\circ + 4\sin120^\circ = 7(\sqrt3/2) \approx 6.06\,\mathrm{cm}\). \(A = \sqrt{X^2 + Y^2} = \sqrt{39} \approx 6.24\,\mathrm{cm}\); \(\phi = \arctan(Y/X) = \arctan(4.04) \approx 76.1^\circ\).