Energy Exchange and Conservation in SHM

Derivation

Energy Exchange and Conservation in SHM

D-058 Home PU-103 Threads energy Depends on Simple Harmonic Motion from a Linear Restoring Force

Statement

For a one-dimensional undamped simple harmonic oscillator of mass \(m\) and stiffness \(k\), executing \(x(t)=A\cos(\omega_0 t+\varphi)\) with \(\omega_0=\sqrt{k/m}\), the total mechanical energy \(E=T+U\) is constant in time and equal to \(\tfrac12 kA^2\), while the kinetic part \(T\) and potential part \(U\) each oscillate about the mean value \(\tfrac14 kA^2\) at angular frequency \(2\omega_0\) — twice the oscillation frequency — exactly out of phase with one another.

Why it matters

This is the mechanical archetype of energy conservation in a conservative bound system: the total is fixed by the amplitude alone, and motion is nothing more than energy sloshing between two reservoirs. Every resonator you meet later — LC circuits, phonon modes, photon field modes, molecular vibrations — inherits this same \(\tfrac12 kA^2\) bookkeeping and the same factor-of-two doubling of the energy-exchange frequency.

Practically, the constant-\(E\) statement lets you solve for the speed at any displacement without ever integrating the equation of motion, and the frequency doubling is the fingerprint experimentalists use to distinguish a genuine energy signal (dissipated power, radiation pressure) from the raw displacement signal.

Assumptions

Linear restoring force only.If any non-Hooke term is present the potential is no longer \(\tfrac12 kx^2\), the motion is anharmonic, and the exchange is no longer purely sinusoidal at the single frequency \(2\omega_0\).

No damping or drag.A velocity-dependent force does net negative work over a cycle, so \(dE/dt<0\) and \(E\) decays rather than staying constant.

No external driving.An applied time-dependent force injects or removes energy, breaking \(dE/dt=0\) even with zero damping.

The stiffness \(k\) and mass \(m\) are time-independent.If \(k(t)\) varies (a parametric oscillator, e.g. a child pumping a swing) the system can gain energy at the pump frequency and \(\tfrac12 kA^2\) is not conserved even without friction.

The zero of potential energy is taken at equilibrium \(x=0\).A shifted reference adds a constant to \(U\) and hence to \(E\); the constant is physically irrelevant but the clean result \(E=\tfrac12 kA^2\) assumes \(U(0)=0\).

Derivation

\[ F(x)=-kx \]

Starting point: the linear restoring force established in the prior result, with \(k>0\). A

\[ U(x)=-\int_0^x F(x')\,dx'=\int_0^x kx'\,dx'=\tfrac12 kx^2 \]

Definition of potential energy for a conservative 1-D force; the force depends on position alone so the line integral is path-independent. Reference \(U(0)=0\). A

\[ x(t)=A\cos(\omega_0 t+\varphi),\qquad v(t)=\dot x=-A\omega_0\sin(\omega_0 t+\varphi) \]

The solution of the equation of motion from the prior result, differentiated once to get the velocity. A

\[ T=\tfrac12 mv^2=\tfrac12 mA^2\omega_0^2\sin^2(\omega_0 t+\varphi) \]

Kinetic energy by definition; substitute \(v\) from step 3 and square. A

\[ m\omega_0^2=m\cdot\frac{k}{m}=k \;\Rightarrow\; T=\tfrac12 kA^2\sin^2(\omega_0 t+\varphi) \]

Use \(\omega_0^2=k/m\) to replace the prefactor; this is what makes \(T\) and \(U\) share the identical amplitude \(\tfrac12 kA^2\). B

\[ U=\tfrac12 kx^2=\tfrac12 kA^2\cos^2(\omega_0 t+\varphi) \]

Substitute the trajectory from step 3 into the potential from step 2. A

\[ E=T+U=\tfrac12 kA^2\left[\sin^2(\omega_0 t+\varphi)+\cos^2(\omega_0 t+\varphi)\right]=\tfrac12 kA^2 \]

Add steps 5 and 6 and apply the Pythagorean identity \(\sin^2+\cos^2=1\). The time dependence cancels exactly — this is the conservation statement. A

\[ \frac{dE}{dt}=mv\dot v+kx\dot x=v(m\ddot x+kx)=0 \]

Independent check without invoking the explicit solution: differentiate \(E=\tfrac12 mv^2+\tfrac12 kx^2\) and use the equation of motion \(m\ddot x+kx=0\). Conservation follows directly from the dynamics for any initial condition. C

\[ \sin^2\theta=\tfrac12\big[1-\cos 2\theta\big],\qquad \cos^2\theta=\tfrac12\big[1+\cos 2\theta\big],\quad \theta=\omega_0 t+\varphi \]

Apply the double-angle identities to steps 5 and 6 to expose the exchange frequency. B

\[ T=\tfrac14 kA^2\big[1-\cos 2\theta\big],\qquad U=\tfrac14 kA^2\big[1+\cos 2\theta\big] \]

Each energy oscillates about the mean \(\tfrac14 kA^2\) with amplitude \(\tfrac14 kA^2\) at angular frequency \(2\omega_0\); the \(\cos 2\theta\) terms are equal and opposite, cancelling in the sum — recovering \(E=\tfrac12 kA^2\) and proving the exchange runs at twice the oscillation frequency, \(180^\circ\) out of phase. C

Result

\[ E=T+U=\tfrac12 kA^2=\tfrac12 m\omega_0^2 A^2=\text{const},\qquad \langle T\rangle=\langle U\rangle=\tfrac14 kA^2,\qquad f_{\text{exch}}=2f_0 \]

Reading. The total energy is set once and for all by the amplitude and stiffness; it does not depend on where in the cycle you look. Kinetic and potential energy each swing between \(0\) and the full \(\tfrac12 kA^2\), perfectly anti-correlated, so that whenever one is maximal the other is zero. Averaged over a cycle they are equal (the virial result for a quadratic potential), each carrying half the energy. Because \(\sin^2\) and \(\cos^2\) complete a full cycle in half the period of \(\sin\) and \(\cos\), the trading happens at \(2\omega_0\) — twice the oscillation frequency.

Units check. \([k][A]^2=(\mathrm{N\,m^{-1}})(\mathrm{m^2})=\mathrm{N\,m}=\mathrm{J}\). Equivalently \([m][\omega_0]^2[A]^2=\mathrm{kg}\cdot\mathrm{s^{-2}}\cdot\mathrm{m^2}=\mathrm{kg\,m^2\,s^{-2}}=\mathrm{J}\). Both forms are energies, as required, and the argument \(2\omega_0 t\) is dimensionless.

Limiting cases

Zero amplitude \((A\to0)\): \(E\to0\) and the mass sits at rest at equilibrium — no energy, no exchange.
Turning points \((x=\pm A)\): \(v=0\) so \(T=0\) and \(U=\tfrac12 kA^2\) — all energy is potential.
Passing through equilibrium \((x=0)\): speed is maximal, \(v_{\max}=A\omega_0\), so \(U=0\) and \(T=\tfrac12 kA^2\) — all energy is kinetic.
Cycle average: \(\langle T\rangle=\langle U\rangle=\tfrac14 kA^2=\tfrac12 E\), the equipartition/virial split for a quadratic potential.
Stiff spring \((k\to\infty\) at fixed \(A)\): \(E\) grows without bound — more energy is stored for the same excursion.

Breaks when

Damping is present. With a drag force \(-bv\), \(dE/dt=-bv^2\le0\): energy leaks monotonically and the amplitude decays, so \(\tfrac12 kA^2\) is no longer constant (\(A\) itself becomes time-dependent).
The potential is anharmonic. If \(U\) contains \(x^3, x^4,\dots\) terms (large-amplitude pendulum, real molecular bonds), \(T\) and \(U\) still sum to a constant but no longer trade sinusoidally at a single frequency \(2\omega_0\) — the exchange acquires harmonics and the clean \(\tfrac14 kA^2\) mean fails.
The oscillator is driven or parametrically pumped. An external force \(F(t)\) does net work over a cycle, and a time-varying \(k(t)\) can feed energy in at \(2\omega_0\); in either case \(E\) is not conserved.
Relativistic or radiative regimes. When \(v\to c\), \(T\neq\tfrac12 mv^2\), or when the oscillator radiates (an accelerating charge), energy is carried away and the closed-system bookkeeping breaks.

Failure modes

Forgetting the \(\omega_0^2=k/m\) substitution. Leaving \(T=\tfrac12 mA^2\omega_0^2\sin^2\) and \(U=\tfrac12 kA^2\cos^2\) with mismatched prefactors, then wrongly concluding the sum is not constant. The prefactors are equal precisely because \(m\omega_0^2=k\).
Claiming \(T\) and \(U\) oscillate at \(\omega_0\). They contain \(\sin^2/\cos^2\), which repeat at \(2\omega_0\). Students read off the frequency of \(x(t)\) instead of the frequency of the energy.
Confusing peak with mean. \(\langle\sin^2\rangle=\langle\cos^2\rangle=\tfrac12\) gives \(\langle T\rangle=\langle U\rangle=\tfrac14 kA^2\), not \(\tfrac12 kA^2\).
Using \(v_{\max}=A\omega_0\) at the wrong point. Maximum speed is at \(x=0\), not at the turning points; plugging \(v_{\max}\) in at \(x=\pm A\) double-counts the energy.
Wrong potential zero. Writing \(U=\tfrac12 k(x-x_0)^2\) but measuring \(x\) from a different origin, so \(U\) is not minimised at the actual equilibrium.
Assuming conservation for a driven or damped system. Applying \(E=\tfrac12 kA^2\) to a decaying or forced oscillation where \(A\) is not constant.

Discussion

The deep content of this result is that a bound conservative system in one dimension has a single conserved scalar — the energy — and that this scalar organises the entire motion. Fixing \(E=\tfrac12 kA^2\) confines the trajectory to a closed curve in the \((x,v)\) phase plane: rescaling \(v\) by \(1/\omega_0\), the constant-energy contour \(v^2/\omega_0^2+x^2=A^2\) is a circle of radius \(A\) traced at uniform angular rate \(\omega_0\). The uniform rate around that circle is why the phase advances linearly and why the period is amplitude-independent (isochronism).

The factor of two in the exchange frequency is not a coincidence but a statement about quadratic quantities. Any bilinear energy density built from a single-frequency amplitude beats at twice that frequency, because squaring a sinusoid produces a DC term plus a component at double the frequency (the identity \(\sin^2\theta=\tfrac12-\tfrac12\cos 2\theta\) is exactly this). The same algebra explains why instantaneous power in AC circuits pulses at \(2f\), why intensity beats at twice the field frequency in optics, and why a mass on a spring pumped by moving its support responds at \(2\omega_0\) — the parametric-resonance condition.

The cycle-averaged equality \(\langle T\rangle=\langle U\rangle\) is the harmonic-oscillator instance of the virial theorem: for a potential \(U\propto x^{n}\), \(\langle T\rangle=\tfrac{n}{2}\langle U\rangle\), and \(n=2\) gives the clean 50/50 split. This equal partition between the kinetic and potential reservoirs is what later becomes the classical equipartition theorem — \(\tfrac12 k_B T\) per quadratic degree of freedom — and it is why a harmonic mode holds \(k_B T\) of thermal energy, not \(\tfrac12 k_B T\).

Structurally, energy conservation here is a consequence of time-translation symmetry: the Lagrangian \(L=\tfrac12 m\dot x^2-\tfrac12 kx^2\) has no explicit \(t\)-dependence, so by Noether's theorem the associated conserved quantity — the Hamiltonian \(H=\dfrac{p^2}{2m}+\tfrac12 kx^2\) — is constant along the motion. Damping and driving break exactly this symmetry (the former via an explicitly velocity-dependent, non-conservative force outside the Lagrangian framework, the latter via explicit time dependence), which is the unified reason both spoil conservation. In the quantised version the same Hamiltonian gives energy levels \(E_n=(n+\tfrac12)\hbar\omega_0\), and the classical \(\tfrac12 kA^2\) is recovered as the coherent-state expectation value in the correspondence limit.

Common misconceptions. Energy is not "used up" as the mass moves — in the ideal case it is only relocated between kinetic and potential form, never lost. The oscillator is not "fastest where the force is strongest": the force is largest at the turning points where the speed is zero, and zero at equilibrium where the speed is greatest. And the energy does not oscillate at the same rate as the position — it beats at double the frequency.

Worked examples

\[ \text{Given }m=0.50\ \mathrm{kg},\ k=200\ \mathrm{N\,m^{-1}},\ A=0.10\ \mathrm{m}. \]

Find \(E\), the speed at \(x=0.060\ \mathrm m\), and the exchange frequency. Set up symbolically first, then insert numbers with units. A

\[ E=\tfrac12 kA^2=\tfrac12(200)(0.10)^2=\tfrac12(200)(0.010)=1.0\ \mathrm{J} \]

Total energy from the boxed result. A

\[ \tfrac12 mv^2=E-\tfrac12 kx^2 \;\Rightarrow\; v=\sqrt{\frac{k}{m}\left(A^2-x^2\right)} \]

Solve conservation for \(v\) at arbitrary \(x\) — no integration of the motion needed. B

\[ v=\sqrt{\frac{200}{0.50}\left(0.10^2-0.060^2\right)}=\sqrt{400\times0.0064}=\sqrt{2.56}=1.6\ \mathrm{m\,s^{-1}} \]

Insert numbers. \(\omega_0=\sqrt{k/m}=\sqrt{400}=20\ \mathrm{rad\,s^{-1}}\), so \(f_0=20/2\pi\approx3.18\ \mathrm{Hz}\). A

\[ E=1.0\ \mathrm{J},\qquad v(0.060\ \mathrm m)=1.6\ \mathrm{m\,s^{-1}},\qquad f_{\text{exch}}=2f_0\approx6.4\ \mathrm{Hz} \]

Check. At \(x=0.060\ \mathrm m\), \(U=\tfrac12(200)(0.060)^2=0.36\ \mathrm J\) and \(T=\tfrac12(0.50)(1.6)^2=0.64\ \mathrm J\); \(T+U=1.0\ \mathrm J\), consistent.

\[ \text{Given a pendulum }L=1.0\ \mathrm m,\ m=0.30\ \mathrm{kg},\ \theta_0=0.15\ \mathrm{rad}. \]

Treating it as SHM, find the total energy, the maximum bob speed, and the kinetic fraction at \(\theta=\theta_0/2\). Small-angle pendulum is SHM with effective stiffness \(k_{\text{eff}}=mg/L\) in the arclength coordinate \(s=L\theta\); amplitude \(A=L\theta_0\). B

\[ E=\tfrac12 k_{\text{eff}}A^2=\tfrac12\left(\frac{mg}{L}\right)(L\theta_0)^2=\tfrac12 mgL\,\theta_0^2 \]

Substitute \(k_{\text{eff}}\) and \(A\); the \(L\) algebra leaves the familiar \(\tfrac12 mgL\theta_0^2\), matching \(mgL(1-\cos\theta_0)\approx\tfrac12 mgL\theta_0^2\). B

\[ E=\tfrac12(0.30)(9.81)(1.0)(0.15)^2=\tfrac12(0.30)(9.81)(0.0225)=0.0331\ \mathrm J\approx33\ \mathrm{mJ} \]

Insert numbers, \(g=9.81\ \mathrm{m\,s^{-2}}\). A

\[ v_{\max}=A\omega_0=L\theta_0\sqrt{\frac{g}{L}}=\theta_0\sqrt{gL}=0.15\sqrt{9.81}=0.47\ \mathrm{m\,s^{-1}} \]

Peak speed at the bottom, where all energy is kinetic: \(\tfrac12 mv_{\max}^2=E\) confirms \(0.47\ \mathrm{m\,s^{-1}}\). A

\[ \frac{T}{E}=1-\frac{U}{E}=1-\left(\frac{\theta}{\theta_0}\right)^2=1-\left(\tfrac12\right)^2=\tfrac34 \]

Since \(U\propto\theta^2\) and equals \(E\) at \(\theta=\theta_0\), the potential fraction is \((\theta/\theta_0)^2\). B

\[ E\approx33\ \mathrm{mJ},\qquad v_{\max}\approx0.47\ \mathrm{m\,s^{-1}},\qquad \frac{T}{E}=\tfrac34\ \text{at}\ \theta=\tfrac{\theta_0}{2} \]

Reading. Halving the angular displacement leaves three-quarters of the energy in kinetic form, because potential energy scales as the square of displacement — a direct consequence of the quadratic well.

Problems

(A) A \(0.20\ \mathrm{kg}\) mass oscillates on a spring with \(k=80\ \mathrm{N\,m^{-1}}\) and amplitude \(5.0\ \mathrm{cm}\). Find the total energy and the maximum speed.
Solution
\(E=\tfrac12 kA^2=\tfrac12(80)(0.050)^2=\tfrac12(80)(0.0025)=0.10\ \mathrm J.\) Maximum speed at \(x=0\): \(\tfrac12 mv_{\max}^2=E\Rightarrow v_{\max}=\sqrt{2E/m}=\sqrt{0.20/0.20}=1.0\ \mathrm{m\,s^{-1}}.\) Check: \(v_{\max}=A\omega_0=0.050\sqrt{80/0.20}=0.050\times20=1.0\ \mathrm{m\,s^{-1}}.\)
(A) At what displacement \(x\) (as a fraction of \(A\)) are the kinetic and potential energies equal?
Solution
Set \(T=U\). Since \(U=\tfrac12 kx^2\) and \(E=\tfrac12 kA^2\), \(T=U\) means \(U=\tfrac12 E\), so \(\tfrac12 kx^2=\tfrac12(\tfrac12 kA^2)\Rightarrow x^2=A^2/2\Rightarrow x=A/\sqrt2\approx0.707A.\) In time, this occurs when \(\cos^2(\omega_0 t+\varphi)=\tfrac12\), i.e. at \(\omega_0 t+\varphi=45^\circ,135^\circ,\dots\) — four times per period.
(B) A mass–spring oscillator has period \(T_p=0.40\ \mathrm s\) and total energy \(0.50\ \mathrm J\). If the mass is \(0.25\ \mathrm{kg}\), find the amplitude.
Solution
\(\omega_0=2\pi/T_p=2\pi/0.40=15.7\ \mathrm{rad\,s^{-1}}.\) Then \(k=m\omega_0^2=0.25\times(15.7)^2=0.25\times246.7=61.7\ \mathrm{N\,m^{-1}}.\) From \(E=\tfrac12 kA^2\): \(A=\sqrt{2E/k}=\sqrt{1.0/61.7}=\sqrt{0.0162}=0.127\ \mathrm m\approx13\ \mathrm{cm}.\) Equivalently \(A=\sqrt{2E/m}/\omega_0=\sqrt{4.0}/15.7=2.0/15.7=0.127\ \mathrm m.\)
(B) Show that the instantaneous rate at which the spring does work on the mass, \(P=-kxv\), oscillates at \(2\omega_0\), and find its amplitude.
Solution
\(P=dT/dt=-dU/dt.\) With \(x=A\cos\theta\), \(v=-A\omega_0\sin\theta\) (\(\theta=\omega_0 t+\varphi\)): \(P=-k(A\cos\theta)(-A\omega_0\sin\theta)=kA^2\omega_0\sin\theta\cos\theta=\tfrac12 kA^2\omega_0\sin 2\theta.\) The \(\sin 2\theta\) factor oscillates at \(2\omega_0\), with amplitude \(\tfrac12 kA^2\omega_0=E\omega_0\). This is the power flowing back and forth between the two reservoirs.
(C) A weakly damped oscillator loses a fraction \(\varepsilon=0.02\) of its energy per cycle. Estimate how many cycles until the energy falls to half its initial value, and comment on whether \(\tfrac12 kA^2\) is still conserved.
Solution
Per cycle \(E\to(1-\varepsilon)E\), so after \(N\) cycles \(E_N=E_0(1-\varepsilon)^N.\) Set \((1-0.02)^N=\tfrac12\): \(N=\ln(\tfrac12)/\ln(0.98)=(-0.693)/(-0.0202)\approx34\) cycles. Energy is conserved only over timescales short compared with this: on a single-cycle scale the undamped result \(\tfrac12 kA^2\) holds to within 2%, but over tens of cycles \(A\) is no longer constant and exact conservation fails. Formally \(E(t)=E_0 e^{-t/\tau}\) with \(\tau=T_p/\varepsilon\) the energy decay time; the quality factor is \(Q=2\pi/\varepsilon\approx314.\)