Energy in SHM

Watch a child on a playground swing at the top of its arc, hanging for an instant almost motionless in the air, and then watch the same swing rocket through the lowest point of its path, a blur of speed. Nothing mysterious is happening — energy is simply changing its address. At the top the swing carries height and almost no speed; at the bottom it carries speed and almost no height. A mass bobbing on a spring, a tuning fork's prong, a guitar string plucked and let go — every simple harmonic oscillator plays out the very same trade, over and over, forever (in the idealised, friction-free world of this chapter). This is the story of energy in SHM: kinetic energy (KE) and potential energy (PE) constantly exchange places as a particle swings between its extreme positions, yet their sum — the total mechanical energy $E$ — never changes. It is one of the cleanest, most testable consequences of the force law you have already met, $F = -kx$ , and it converts a motion problem into a bookkeeping problem: energy in one form must reappear as the other, because nothing has anywhere else to go. This section builds directly on the displacement, velocity and force-law results already derived for SHM. Using nothing more than $v = \omega\sqrt{A^2-x^2}$ and $F=-kx$ , we will derive clean, exam-ready expressions for KE and PE at any displacement $x$ , show algebraically that their sum is a constant equal to $\tfrac12 m\omega^2A^2$ , and use that single conserved quantity to shortcut problems that would otherwise need kinematics. Along the way we will look at the KE-versus- $x$ and PE-versus- $x$ graphs, work two numerical examples (one board-style, one calculator-speed competitive), and flag the traps that most reliably cost marks in this topic.

Key takeaways

A particle in SHM continuously exchanges energy between kinetic energy (KE) and potential energy (PE), but the sum $E = KE + PE$ stays constant throughout the motion.
$KE = \tfrac12 m\omega^2(A^2-x^2)$ — maximum at the mean position ( $x=0$ ), zero at the extremes ( $x=\pm A$ ).
$PE = \tfrac12 m\omega^2 x^2$ — zero at the mean position, maximum at the extremes.
Total energy $E = \tfrac12 m\omega^2 A^2 = \tfrac12 kA^2$ (using $\omega^2 = k/m$ ) — independent of $x$ and of time.
$E$ scales as $A^2$ : doubling the amplitude quadruples the energy, not doubles it.
The KE-versus- $x$ and PE-versus- $x$ graphs are mirror-image parabolas that always add up to the same constant $E$ — a fast visual check on any calculation.
Key idea: as an SHM particle moves, kinetic energy $\tfrac12 m\omega^2(A^2-x^2)$ and potential energy $\tfrac12 m\omega^2 x^2$ continuously trade places, but their sum stays fixed at $E=\tfrac12 m\omega^2A^2$ , letting you find speed or displacement from energy conservation alone.

The core insight

In SHM, kinetic energy $\tfrac12 m\omega^2(A^2-x^2)$ and potential energy $\tfrac12 m\omega^2 x^2$ continuously convert into each other as the particle moves, but their sum $E=\tfrac12 m\omega^2A^2$ stays constant at every instant and every position.

Where it fits in the syllabus

Energy in SHM sits right after the displacement, velocity/acceleration and force-law sections of the CBSE Class 11 Physics chapter "Oscillations," and it is explicitly on the current 2025-26 prescription: you are expected to derive $KE = \tfrac12 m\omega^2(A^2-x^2)$ , $PE = \tfrac12 m\omega^2x^2$ , and show $E = \tfrac12 m\omega^2A^2$ is constant. This is one of the most reliably tested application topics in the chapter, because it needs only the velocity formula you already know plus a small piece of algebra. This section stays strictly within that scope — it does not touch damped or forced oscillations, which involve energy loss and belong to a later stage of study (Class 12 and beyond).

The physical picture

Picture a particle of mass $m$ executing SHM about a mean position $O$ , with amplitude $A$ and angular frequency $\omega$ . At any instant its displacement from $O$ is $x$ , and — from the velocity relation you derived earlier — its speed there is $v = \omega\sqrt{A^2-x^2}$ . Kinetic energy is just $\tfrac12mv^2$ , so substituting this speed gives the kinetic energy at that instant purely in terms of $x$ :

KE = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2\left(A^2-x^2\right)

where: $m$ is the oscillating mass (kg), $\omega$ is the angular frequency (rad/s), $A$ is the amplitude (m), and $x$ is the instantaneous displacement (m) from the mean position. Physically, $KE$ is largest exactly where speed is largest — the mean position, $x=0$ — and shrinks to zero at the extreme positions $x=\pm A$ , where the particle momentarily stops to reverse direction. The restoring force $F=-kx$ does work on the particle as it moves, and that work is stored as potential energy. Taking $PE=0$ at the mean position (the natural, unstretched reference for a spring), the potential energy stored at displacement $x$ works out to the familiar elastic-PE form, now written using $\omega^2=k/m$ :

PE = \frac{1}{2}m\omega^2 x^2 \left(= \frac{1}{2}kx^2\right)

where: the symbols are as above; note $PE$ grows purely as $x^2$ , so it is zero at the mean position and largest at the extremes, exactly the opposite pattern to $KE$ . Adding the two expressions gives the total mechanical energy, and the $x^2$ terms cancel perfectly, leaving a result with no $x$ in it at all:

E = KE + PE = \frac{1}{2}m\omega^2\left(A^2-x^2\right) + \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m\omega^2A^2

where: $E$ is the total mechanical energy (J), fixed once $m$ , $\omega$ and $A$ are fixed for a given oscillator. Since it does not depend on $x$ , $E$ is the same at every point of the motion and at every instant of time — this is conservation of mechanical energy applied to an ideal, friction-free oscillator, and it is the single most useful fact in this whole topic.

Figure

Two mirror-image parabolas plotted against displacement from minus A to plus A, one for kinetic energy peaking at the centre and falling to zero at both ends, the other for potential energy starting at zero in the centre and rising to a peak at both ends, with their sum shown as a flat horizontal line equal to the total energy

It is worth reading the graph as carefully as the algebra. The $PE$ -versus- $x$ curve is an upward-opening parabola, zero at the centre and rising symmetrically to a maximum at both extremes; the $KE$ -versus- $x$ curve is a downward-opening parabola, a mirror image of it. Because the two are complementary — whatever $PE$ loses, $KE$ gains, and vice versa — their sum traces a flat horizontal line at height $E$ . This gives a fast sanity check for any numerical problem: compute one energy at a given $x$ , subtract from the known total $E$ , and the other energy falls out for free, with no need to touch the velocity formula again.

Derivation ATotal mechanical energy in SHM is constant

The goal of this derivation is to start from the two separate expressions for kinetic and potential energy of a particle executing SHM, each written as a function of the instantaneous displacement $x$ , and to show algebraically that their sum collapses to a single constant that depends only on the fixed properties of the oscillator — mass, angular frequency and amplitude — and not on $x$ or time at all, which is the precise statement of energy conservation for this motion.

Step 1. Start from the SHM velocity relation derived earlier in the chapter, $v = \omega\sqrt{A^2-x^2}$ , obtained from $x=A\cos(\omega t+\phi)$ and $v=-A\omega\sin(\omega t+\phi)$ by eliminating time using $\sin^2+\cos^2=1$ . Kinetic energy is $KE=\tfrac12mv^2$ , so squaring $v$ and substituting gives $KE = \tfrac12 m\omega^2(A^2-x^2)$ , a downward parabola in $x$ with maximum value $\tfrac12m\omega^2A^2$ at $x=0$ and value zero at $x=\pm A$ .

Step 2. Find the potential energy by computing the work done against the restoring force $F=-kx$ as the particle moves from the mean position out to displacement $x$ . The work done by an external agent against this force, stored as PE, is $PE = \int_0^x kx'\,dx' = \tfrac12kx^2$ . Since $k=m\omega^2$ (from the force-law relation $\omega=\sqrt{k/m}$ derived earlier), this becomes $PE = \tfrac12m\omega^2x^2$ , an upward parabola in $x$ , zero at $x=0$ and maximum at $x=\pm A$ .

Step 3. Add the two expressions term by term. $E = KE+PE = \tfrac12m\omega^2(A^2-x^2) + \tfrac12m\omega^2x^2$ . Expand the bracket: $\tfrac12m\omega^2A^2 - \tfrac12m\omega^2x^2 + \tfrac12m\omega^2x^2$ . The two $x^2$ terms are equal in magnitude and opposite in sign, so they cancel exactly, leaving $E = \tfrac12m\omega^2A^2$ .

Step 4. Read the physical meaning of the result. The final expression has no $x$ and no $t$ in it — only the fixed constants $m$ , $\omega$ and $A$ of a given oscillator. This means the total mechanical energy is the same at every position along the path and at every instant of time; it is set once and for all by the amplitude and angular frequency at the start of the motion. Physically this is exactly what conservation of mechanical energy predicts, because the only force acting in ideal SHM is the conservative restoring force $F=-kx$ — there is no friction or resistance to drain energy away, so whatever KE is lost as the particle slows down going outward is gained back in full as PE, and the reverse happens coming back in.

Worked examples

Board level - energy split of an oscillating block

A particle of mass 0.2 kg executes SHM with amplitude 0.1 m and angular frequency 5 rad/s. Find (a) the total energy, and (b) the KE and PE when the displacement is 0.05 m.

Step 1. Compute the total energy using $E=\tfrac12m\omega^2A^2$ . With $m=0.2\ \text{kg}$ , $\omega=5\ \text{rad/s}$ , $A=0.1\ \text{m}$ : $E = \tfrac12(0.2)(5)^2(0.1)^2 = \tfrac12(0.2)(25)(0.01) = \tfrac12(0.05) = \mathbf{0.025\ J}$ .

Step 2. Compute the potential energy at $x=0.05\ \text{m}$ using $PE=\tfrac12m\omega^2x^2$ : $PE = \tfrac12(0.2)(25)(0.05)^2 = \tfrac12(0.2)(25)(0.0025) = \tfrac12(0.0125) = \mathbf{0.00625\ J}$ .

Step 3. Get the kinetic energy the fast way — by subtraction from the known total, rather than recomputing from the velocity formula. Since $E=KE+PE$ is constant, $KE = E - PE = 0.025 - 0.00625 = \mathbf{0.01875\ J}$ . As a check, the direct formula gives the same value: $KE=\tfrac12m\omega^2(A^2-x^2)=\tfrac12(0.2)(25)(0.01-0.0025)=\tfrac12(0.2)(25)(0.0075)=0.01875\ \text{J}$ , confirming the subtraction shortcut.

Competitive level - speed at a given fraction of amplitude

A block of mass 1 kg attached to a spring of force constant 100 N/m executes SHM with amplitude 0.2 m. Find the speed of the block when it is halfway between the mean position and an extreme position, i.e. at $x=A/2$ .

Step 1. Find $\omega$ from the spring relation, $\omega=\sqrt{k/m}=\sqrt{100/1}=10\ \text{rad/s}$ . This is the quantity that lets every later step avoid dealing with $k$ and $m$ separately.

Step 2. Locate the point: $x=A/2=0.2/2=0.1\ \text{m}$ . Rather than reaching for the standalone velocity formula, use energy conservation directly: $\tfrac12mv^2 = E - PE = \tfrac12m\omega^2A^2 - \tfrac12m\omega^2x^2 = \tfrac12m\omega^2(A^2-x^2)$ , which on cancelling the common factor $\tfrac12m$ gives $v=\omega\sqrt{A^2-x^2}$ — the same relation as the kinematic route, reached here purely from energy bookkeeping.

Step 3. Substitute the numbers: $v = 10\sqrt{(0.2)^2-(0.1)^2} = 10\sqrt{0.04-0.01} = 10\sqrt{0.03} \approx 10\times0.1732 \approx \mathbf{1.73\ m/s}$ . Timing note: recognise instantly that "speed at $x=A/2$ " is an energy-conservation question, not a fresh kinematics derivation — writing $v=\omega\sqrt{A^2-x^2}$ straight from memory and plugging in numbers should take under 30 seconds; do not waste competitive-exam time re-deriving $KE=E-PE$ from scratch each time.

Common traps

Assuming KE is maximum at the extreme positions — it is the opposite: KE is maximum at the mean position ( $x=0$ ), where speed is highest, and zero at the extremes, where the particle momentarily stops.
Forgetting that total energy $E$ scales as $A^2$ , not $A$ — doubling the amplitude quadruples the total energy; it does not merely double it.
Using $\tfrac12kx^2$ for PE with the wrong reference point — the formula $PE=\tfrac12m\omega^2x^2=\tfrac12kx^2$ only holds when $x$ is measured from the mean (equilibrium) position, since that is where PE is defined to be zero in this chapter's convention.
Recomputing "new" total energy at every position instead of realising $E=KE+PE$ never changes — many mark losses come from redoing the full velocity-and-KE calculation when a simple subtraction from the known constant $E$ would have worked.
Mixing up which quantity is asked for — a question giving $x$ and asking for speed wants the velocity formula (or energy conservation solved for $v$ ); a question giving $v$ and asking for $x$ needs the same relation solved the other way round; read carefully before choosing a formula.

You should be able to…

Derive $KE=\tfrac12m\omega^2(A^2-x^2)$ and $PE=\tfrac12m\omega^2x^2$ for a particle in SHM, starting from the velocity relation and the restoring force.
Show algebraically that $E=KE+PE=\tfrac12m\omega^2A^2$ is independent of $x$ and of time.
State where KE and PE are each maximum and minimum during one oscillation, and sketch their graphs against $x$ .
Use energy conservation ( $KE=E-PE$ ) to find speed or displacement at a given point, without redoing the full kinematic derivation each time.
Relate the SHM energy expressions to the spring's force constant, $E=\tfrac12kA^2$ , using $\omega^2=k/m$ .
Recognise that constant $E$ assumes an ideal, friction-free oscillator, and that real oscillators lose energy over time (damped oscillations — outside this chapter).

Ask the tutor: Want a visual walkthrough of how KE and PE trade off across a full oscillation, or more practice problems combining energy conservation with the spring and pendulum formulas — ask the AI tutor to guide you through it.

Why is PE taken as zero at the mean position and not at the extreme position?

It is a choice of reference, not a law of physics — potential energy is always measured relative to some chosen zero. In SHM the natural, universally-used convention is to set PE = 0 at the mean (equilibrium) position, because that is where the restoring force is zero and the spring or string is at its natural configuration. With this choice, PE grows purely as $x^2$ outward from the centre, which is exactly what makes $PE=\tfrac12m\omega^2x^2=\tfrac12kx^2$ hold. If you shifted the zero elsewhere, the formula would need an extra constant term, so always assume the mean-position convention unless a question states otherwise.

Why does the total energy depend on A squared and not just A?

Both KE and PE, when written as functions of $x$ , come from squaring a length — velocity squared in $KE=\tfrac12mv^2$ , and displacement squared in $PE=\tfrac12kx^2$ . Since $v=\omega\sqrt{A^2-x^2}$ also has $A$ appearing inside a square root that gets squared back when you compute $KE$ , every term in the energy expressions ends up proportional to $A^2$ , never to $A$ alone. So doubling the amplitude does not double how far the particle swings and proportionally double the energy — it quadruples the peak speed-squared and peak displacement-squared simultaneously, and the energy follows suit.

Can I use energy conservation instead of the v = omega root A squared minus x squared formula in every problem?

Yes — they are exactly the same physics stated two ways. Energy conservation, $\tfrac12mv^2 = E-PE = \tfrac12m\omega^2A^2-\tfrac12m\omega^2x^2$ , simplifies algebraically to the identical velocity formula once you cancel $\tfrac12m$ and take a square root. Many students find it faster in practice to think in terms of "total energy minus potential energy gives kinetic energy, then solve for v" because it reuses the same E you may have already calculated in an earlier part of the question, avoiding a second independent calculation.

Does the block ever actually stop moving during SHM?

Yes, but only for an instant, at each extreme position $x=\pm A$ . At that instant $v=0$ , so $KE=0$ and all the energy is stored as $PE=\tfrac12m\omega^2A^2=E$ . The restoring force is at its maximum there too (since $F=-kx$ is largest in magnitude at $x=\pm A$ ), which is exactly why the particle does not stay at rest — it is immediately accelerated back toward the mean position, and the whole energy exchange repeats in reverse on the way back.