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17.2 Energy in simple harmonic motion

Syllabus
9702–2028–2029
Topic
17.2
Level
A2

Kinetic and potential energy interchange during ideal SHM

In ideal undamped SHM, kinetic energy E_K and potential energy E_P continually interchange while total mechanical energy E=E_K+E_P remains constant.

Position Speed Kinetic energy Potential energy
x=0, equilibrium maximum maximum, E minimum, taken as 0
0< x <x0 between 0 and maximum
x=±x0, extremes 0 0 maximum, E

EP=(1/2)mω2x2EK=(1/2)mω2(x02−x2)E=(1/2)mω2x02E_P=(1/2)mω²x² E_K=(1/2)mω²(x0²-x²) E=(1/2)mω²x0²

Against x, E_P is an upward U-shaped parabola with minimum at x=0; E_K is a downward arch, zero at ±x0 and maximum at x=0; total E is a horizontal line.

Each energy is the same at +x and -x and reaches two maxima per oscillation, so energy-time graphs repeat with period T/2.

Energy is not created at equilibrium. It has changed form. With damping, mechanical energy decreases; with driving, energy may enter, so the ideal constant-total-energy statement needs that boundary.

Total energy in ideal SHM is one-half m omega-squared amplitude-squared

E=(1/2)mω2x02E=(1/2)mω²x0²

Here m is the oscillating mass, ω is angular frequency and x0 is amplitude. In ideal SHM, E equals maximum kinetic energy at equilibrium and maximum potential energy at either extreme.

Atx=0,v=v0=ωx0,soE=EK,max=(1/2)mv02=(1/2)mω2x02.At x=0, v=v0=ωx0, so E=E_K,max=(1/2)mv0²=(1/2)mω²x0².

For m=0.150 kg, ω=15.7 rad s⁻¹ and x0=0.0160 m, E=(1/2)(0.150)(15.7)²(0.0160)²=4.73×10⁻³ J.

If an oscillator travels 14 mm in one complete cycle, it covers 4x0, so x0=3.5 mm. Then ω=sqrt(2E/(mx0²)) when E and m are known.

E∝m,E∝ω2,E∝x02;doublingamplitudequadruplesE.E ∝ m, E ∝ ω², E ∝ x0²; doubling amplitude quadruples E.

Use amplitude, not instantaneous displacement, peak-to-peak distance or total distance per cycle. Convert all lengths to metres before substitution.