Physics · Ch 10 — Oscillations

ENERGY IN SIMPLE HARMONIC MOTION

10.5

ENERGY IN SIMPLE HARMONIC MOTION

As a particle executes SHM, it continually exchanges energy between two forms -- kinetic energy (KE), associated with its speed, and potential energy (PE, or elastic PE for a spring), associated with its displacement from the mean position -- while its total mechanical energy stays exactly constant throughout the motion (in the idealised, undamped case). The next three parts derive the PE, KE, and total energy expressions in turn, and show explicitly that their sum is a time-i …

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Expression for Potential Energy

The SHM restoring force obeys Hooke's law, $F=-kx$ (the one-dimensional case of $\vec F=-k\vec r$ ), and is conservative, so it can be derived from a scalar potential energy function via $F=-dU/dx$ . Comparing the two gives $dU=kx\,dx$ ; integrating from the mean position ( $x'=0$ , $U=0$ ) out to displacement $x$ , $U(x)=\int_0^x kx'\,dx'=\tfrac12 kx^2$ (the integration variable $x'$ is a dummy variable -- any symbol gives the same result). Substituting the SHM relation $k=m\omega^2$ turns this into $U(x)=\tfrac12 m\omega^2x^2$ . For a particle actually executing SHM, $x=A\sin\omega t$ , so the potential energy varies with time as $U(t)=\tfrac12 m\omeg …

Figure 10.23Variation of potential energy with time

What this figure shows. A curve of potential energy U(t) plotted against time t over one full oscillation period T shows U rising from zero (at the mean position, t = 0) to a maximum (at the extreme position, t = T/4), back down to zero (mean position again, t = T/2), up to the same maximum again (opposite extreme, t = 3T/4), and back to zero at t = T. Because U depends on sin-squared of the phase, the curve completes two full up-down humps within one period T, i.e. potential energy itself oscillates with period T/2, tw …

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Expression for Kinetic Energy

Kinetic energy is $KE=\tfrac12 mv^2$ , where $v=dx/dt$ . For a particle in SHM, $x=A\sin\omega t$ gives $v=A\omega\cos\omega t$ , which (using $\cos\omega t=\sqrt{1-\sin^2\omega t}$ and $\sin\omega t=x/A$ ) can be rewritten purely in terms of displacement as $v=\omega\sqrt{A^2-x^2}$ . Substituting into the kinetic-energy formula gives $KE=\tfrac12 m\omega^2(A^2-x^2)$ , or equivalently, using $x=A\sin\omega t$ directly, $KE=\tfrac12 m\omega^2A^2\cos^2\omega t$ . This is maximum ( $\tfrac12 m\omega^2A^2$ ) exactly at the mean position ( $x=0$ ) and zero at the extreme positions ( $x=\pm A$ ) -- the mirror image of how PE behaves; both $KE(t)$ and $U(t)$ individually osc …

Figure 10.24Variation of kinetic energy with time

What this figure shows. A curve of kinetic energy KE(t) plotted against time t over one full period T shows the mirror-image behaviour to the potential-energy curve: KE starts at its maximum value at t = 0 (the mean position, where speed is greatest), falls to zero at t = T/4 (the extreme position, where the particle is momentarily at rest), rises back to maximum at t = T/2, falls to zero again at t = 3T/4, and returns to maximum at t = T -- also completing two full humps within one period T, so that at every instant where the potential-energy curve is at its peak, …

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Expression for Total Energy

Total mechanical energy is the sum $E=KE+U=\tfrac12 m\omega^2(A^2-x^2)+\tfrac12 m\omega^2x^2$ ; the $x^2$ terms cancel exactly, leaving $E=\tfrac12 m\omega^2A^2=\text{constant}$ -- independent of both time and instantaneous position. The same result follows by adding the time-domain expressions, since $\sin^2\omega t+\cos^2\omega t=1$ : $E=\tfrac12 m\omega^2A^2(\sin^2\omega t+\cos^2\omega t)=\tfrac12 m\omega^2A^2$ . This is the law of conservation of total energy for SHM: while KE and PE individually rise and fall (each periodic, with period $T/2$ ), their sum never changes. Because KE is proportional to $v^2$ it can never be negative, and similarly PE is never negative in SHM; at the mean position the energy is purely kinetic, and at the extreme positions it is purely potential. The total energy gives the amplitude directly: $A=\sqrt{2E/(m\omega^2)}=\sqrt{2E/k}$ . The position at which KE and PE are exactly equa …

Figure 10.25Both kinetic and potential energy vary but total energy is constant

What this figure shows. The kinetic-energy curve K(t) and potential-energy curve U(t) are plotted together on the same axes over one period T, each rising and falling in a sin-squared/cos-squared pattern with period T/2, but always exactly out of step with each other so that at every instant their sum U(t)+K(t) traces a perfectly flat horizontal line at the constant height E -- the total mechanical energy -- visually proving that whatever energy potential energy loses at any moment, kinetic energy gains, and vice versa, keeping t …

Figure 10.26Conservation of energy -- spring-mass system and simple pendulum, energy bar charts

What this figure shows. A set of energy bar charts for a mass on a spring (positions A through E across one swing) and a matching set for a simple pendulum (at the equilibrium point and at each extreme) shows, for each position, a pair of stacked bars for kinetic energy (KE) and potential energy (PE) that always add up to the same total-energy (TE) bar height. At the equilibrium/mean position the KE bar is full height and the PE bar is empty (maximum speed, minimum stored energy); at each extreme position the PE bar is full height and the KE bar is empty (zero speed, maximum stored energy); at in-between positions both bars are partially filled but their sum, the TE b …

Misc Example 10.15Kinetic energy and total energy in terms of momentum

Worked out. The task is to re-express kinetic energy and total energy for one-dimensional SHM in terms of the particle's linear momentum p_x instead of its velocity. Starting from KE = (1/2) m v_x^2, multiplying numerator and denominator by m gives KE = (m v_x)^2/(2m) = p_x^2/(2m), the standard momentum form of kinetic energy. Adding the potential energy term U(x) = (1/2) m omega^2 x^2 gives the total energy as E = p_x^2/(2m) + (1/2) m omega^2 x^2 = constant, and substituting the SHM time dependence and using sin^2 + cos^2 = 1 confirms this constant equals (1/2) m omega^2 A^2, exactly the same total-energy result reached earlier directly in terms of velocity -- showing the momentum form and the velocity form of the energy …

Misc Example 10.16Position where kinetic energy equals potential energy

Worked out. The task is to find the position of an oscillating particle at which its kinetic energy and potential energy are exactly equal. Setting KE = U, i.e. (1/2) m omega^2 (A^2-x^2) = (1/2) m omega^2 x^2, the common factor (1/2) m omega^2 cancels from both sides, leaving A^2 - x^2 = x^2, so 2x^2 = A^2, giving x = +- A/sqrt(2). This position, at about 70.7% of the full amplitude, is where the oscillator's total energy is split exactly evenly between its kinetic and potential forms -- closer to the mean position than to the extreme, since kinetic energy (which is largest at the mean position) has more 'room' to fall befo …

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