Two purses, one fixed total
In simple harmonic motion the particle moves under the restoring force $F=-kx$, where $k=m\omega^2$ (this identity comes straight from the force law for SHM). That force is conservative, so the work it does is fully recoverable and the system carries a well-defined potential energy. As the particle sweeps from a turning point through the centre and out to the other turning point, it trades potential energy for kinetic and back again, while the sum stays locked.
NCERT opens §13.7 with the governing observation: "Both kinetic and potential energies of a particle in SHM vary between zero and their maximum values." The two energies are out of step. Where one is largest the other vanishes, and vice versa. The whole of this topic is the bookkeeping of that exchange.
Kinetic energy in SHM
For a particle of mass $m$ with velocity $v$, the kinetic energy is $K=\tfrac12 mv^2$. The SHM velocity (derived in velocity and acceleration in SHM) is $v=-\omega A\sin(\omega t+\varphi)$, and the velocity-position relation is $v=\pm\,\omega\sqrt{A^2-x^2}$. Substituting either form gives the kinetic energy in three equivalent guises:
$$K=\tfrac12 mv^2=\tfrac12 m\omega^2(A^2-x^2)=\tfrac12 m\omega^2 A^2\sin^2(\omega t+\varphi)$$Because $v^2$ is even in $v$, the sign of velocity does not matter to $K$ — the particle has the same kinetic energy whether moving left or right through a given point. NCERT notes the consequence directly: "since the sign of $v$ is immaterial in $K$, the period of $K$ is $T/2$." $K$ is largest at the mean position ($x=0$, where $\sin^2=1$) and zero at the extremes ($x=\pm A$).
Potential energy in SHM
The potential energy stored in the deformation is the work done against the restoring force in moving from the mean position to displacement $x$. NIOS computes this as the area under the $F=ky$ straight-line graph — the area of a triangle of base $x$ and height $kx$ — giving $U=\tfrac12 kx^2$. Writing $k=m\omega^2$ and inserting $x=A\cos(\omega t+\varphi)$ yields the three standard forms:
$$U=\tfrac12 kx^2=\tfrac12 m\omega^2 x^2=\tfrac12 m\omega^2 A^2\cos^2(\omega t+\varphi)$$This mirrors the kinetic energy exactly. $U$ is zero at the mean position and maximum at the extremes, again repeating with period $T/2$. The potential energy is taken positive by the conventional choice of the additive constant, so on any graph both energies sit above the axis.
Total energy is constant
Add the two. Using $\sin^2\theta+\cos^2\theta=1$ with $\theta=\omega t+\varphi$:
$$E=K+U=\tfrac12 m\omega^2 A^2\big[\sin^2(\omega t+\varphi)+\cos^2(\omega t+\varphi)\big]=\tfrac12 m\omega^2 A^2=\tfrac12 kA^2$$The bracket collapses to unity, leaving an expression with no $t$ and no $x$. NCERT states the meaning plainly: "The total mechanical energy of a harmonic oscillator is thus independent of time as expected for motion under any conservative force." The same constant can be read off two ways — at $x=0$ all of it is kinetic ($E=\tfrac12 m v_{\max}^2=\tfrac12 m\omega^2 A^2$), and at $x=\pm A$ all of it is potential ($E=\tfrac12 kA^2$).
Master comparison: KE vs PE vs E
Every property a NEET item can ask about — formula, location of the maximum, period, time-average — fits in a single grid. The three forms in each cell are interconvertible through $k=m\omega^2$ and $v_{\max}=\omega A$.
| Property | Kinetic energy $K$ | Potential energy $U$ | Total energy $E$ |
|---|---|---|---|
| General formula | $\tfrac12 mv^2$ | $\tfrac12 kx^2=\tfrac12 m\omega^2 x^2$ | $\tfrac12 kA^2=\tfrac12 m\omega^2 A^2$ |
| In terms of $x$ | $\tfrac12 m\omega^2(A^2-x^2)$ | $\tfrac12 m\omega^2 x^2$ | $\tfrac12 m\omega^2 A^2$ (independent of $x$) |
| In terms of $t$ ($x=A\cos$) | $\tfrac12 m\omega^2 A^2\sin^2(\omega t+\varphi)$ | $\tfrac12 m\omega^2 A^2\cos^2(\omega t+\varphi)$ | $\tfrac12 m\omega^2 A^2$ (independent of $t$) |
| Maximum value | $\tfrac12 kA^2$ at $x=0$ (mean) | $\tfrac12 kA^2$ at $x=\pm A$ (extreme) | $\tfrac12 kA^2$ everywhere |
| Minimum value | $0$ at $x=\pm A$ | $0$ at $x=0$ | never changes |
| Period of variation | $T/2$ | $T/2$ | constant (does not oscillate) |
| Frequency of variation | $2n$ (twice the displacement) | $2n$ (twice the displacement) | — |
| Time-average over a cycle | $\tfrac14 kA^2=E/2$ | $\tfrac14 kA^2=E/2$ | $\tfrac12 kA^2$ |
The energy-vs-displacement parabolas
As functions of $x$, both energies are parabolas. $U=\tfrac12 kx^2$ is an upward parabola through the origin; $K=\tfrac12 k(A^2-x^2)$ is a downward parabola peaking at the centre. They are reflections of each other in the horizontal line $E/2$, and at every $x$ they sum to the flat line $E=\tfrac12 kA^2$.
Figure 1. Energy versus displacement. The PE parabola opens upward (zero at the mean, maximum at the extremes); the KE parabola opens downward (maximum at the mean, zero at the extremes). They cross at $x=\pm A/\sqrt2$, where each equals $E/2$. At every $x$ the two add to the constant total $E=\tfrac12 kA^2$.
The crossing point deserves a note. Setting $K=U$ gives $A^2-x^2=x^2$, hence $x=\pm A/\sqrt2\approx\pm0.707A$. Only there is the energy split exactly even at a single instant. Everywhere else the instantaneous division is unequal even though the cycle-average division is precisely half and half.
The $v=\pm\omega\sqrt{A^2-x^2}$ relation that makes $K=\tfrac12 m\omega^2(A^2-x^2)$ comes from velocity and acceleration in SHM — review it before energy problems.
Why energy oscillates at double the frequency
The single most examined subtlety in this topic is the frequency at which the energies vary. Take the time forms and apply the power-reduction identities:
$$K=\tfrac12 m\omega^2 A^2\sin^2(\omega t+\varphi)=\tfrac14 m\omega^2 A^2\big[1-\cos 2(\omega t+\varphi)\big]$$ $$U=\tfrac12 m\omega^2 A^2\cos^2(\omega t+\varphi)=\tfrac14 m\omega^2 A^2\big[1+\cos 2(\omega t+\varphi)\big]$$The oscillating part of each carries $\cos 2(\omega t+\varphi)$ — an angular frequency of $2\omega$, not $\omega$. So while the displacement completes one cycle in period $T$, each energy completes two cycles in the same time, with period $T/2$ and frequency $2n$. NCERT puts it as: "Both kinetic energy and potential energy peak twice during each period of SHM."
Figure 2. Displacement (top) versus the two energies (bottom) on the same time axis. While $x(t)$ runs through one period $T$, KE and PE each run through two — period $T/2$, frequency $2n$. KE and PE are perfectly out of phase, and their sum is the flat total-energy line.
Energy depends on A² and ω²
The total energy $E=\tfrac12 m\omega^2 A^2$ scales as the square of two quantities at once. Doubling the amplitude multiplies the energy by four; doubling the angular frequency does the same. Mass enters linearly. These dependences are summarised below.
| Change made | Effect on total energy $E=\tfrac12 m\omega^2 A^2$ |
|---|---|
| Amplitude $A \to 2A$ (same $\omega$, $m$) | $E \to 4E$ (since $E\propto A^2$) |
| Angular frequency $\omega \to 2\omega$ (same $A$, $m$) | $E \to 4E$ (since $E\propto \omega^2$) |
| Mass $m \to 2m$ (same $A$, $\omega$) | $E \to 2E$ (since $E\propto m$) |
| Frequency halved, amplitude doubled | $E$ unchanged ($\omega^2$ down ×4, $A^2$ up ×4) |
For a spring-mass system $\omega^2=k/m$, so $E=\tfrac12 m\cdot\dfrac{k}{m}\cdot A^2=\tfrac12 kA^2$ and the mass formally cancels — the energy is fixed entirely by the spring constant and the amplitude. Build the geometry of such systems from the spring-mass system note.
Time-averaged energies
Over one full cycle the average of $\sin^2$ and of $\cos^2$ is each $\tfrac12$. Hence the time-averaged kinetic and potential energies are equal, and each is half the total:
$$\langle K\rangle=\langle U\rangle=\tfrac12\cdot\tfrac12 kA^2=\tfrac14 kA^2=\frac{E}{2}$$So although at any chosen instant the energy is generally lopsided — all kinetic at the centre, all potential at the turning points — averaged across the motion it splits exactly in two. This even split is a hallmark of the harmonic potential and a tidy way to remember that $\langle K\rangle=\langle U\rangle$ for SHM.
Worked example — NCERT 13.7
A block of mass 1 kg is fastened to a spring of force constant $k=50~\text{N m}^{-1}$. It is pulled to $x=10$ cm from the equilibrium position and released from rest at $t=0$ on a frictionless surface. Find the kinetic, potential and total energies when the block is 5 cm from the mean position.
Angular frequency. $\omega=\sqrt{k/m}=\sqrt{50/1}=7.07~\text{rad s}^{-1}$. With release from rest at the extreme, $x(t)=0.1\cos(7.07t)$ and $A=0.1$ m.
Total energy (use the amplitude). $E=\tfrac12 kA^2=\tfrac12(50)(0.1)^2=0.25~\text{J}$. This is fixed for the whole motion.
Potential energy at $x=0.05$ m (use the instantaneous displacement). $U=\tfrac12 kx^2=\tfrac12(50)(0.05)^2=0.0625~\text{J}$.
Kinetic energy at $x=0.05$ m. Either $K=\tfrac12 m\omega^2(A^2-x^2)=\tfrac12(1)(50)(0.1^2-0.05^2)=0.1875\approx0.19~\text{J}$, or simply $K=E-U=0.25-0.0625=0.1875~\text{J}$.
Check. $K+U=0.1875+0.0625=0.25~\text{J}=E$, in conformity with conservation of energy. Note also that at $x=0.05=A/2$, the PE is one-quarter of the total ($\tfrac14 kA^2$ scaled by $(x/A)^2$), confirming PE rises as $x^2$.
Energy in SHM in one breath
- $K=\tfrac12 mv^2=\tfrac12 m\omega^2(A^2-x^2)=\tfrac12 m\omega^2 A^2\sin^2(\omega t+\varphi)$ — maximum at the mean position, zero at the extremes.
- $U=\tfrac12 kx^2=\tfrac12 m\omega^2 x^2=\tfrac12 m\omega^2 A^2\cos^2(\omega t+\varphi)$ — zero at the mean position, maximum at the extremes.
- $E=K+U=\tfrac12 kA^2=\tfrac12 m\omega^2 A^2$ — independent of both time and position.
- $E\propto A^2$, $E\propto\omega^2$, $E\propto m$: doubling $A$ or $\omega$ quadruples the energy.
- Both energies oscillate at frequency $2n$ (period $T/2$) — twice the displacement frequency.
- Plotted against $x$, KE and PE are mirror-image parabolas that sum to the flat line $E$; they cross at $x=\pm A/\sqrt2$.
- Time-averaged over a cycle, $\langle K\rangle=\langle U\rangle=\tfrac14 kA^2=E/2$.