12.4 More About Energy in SHM

Derive compact quantitative formulas linking displacement, velocity, acceleration, and energy so mixed-data SHM problems can be solved quickly.

Estimated time: 34 minutes

From x(t) and v(t) to Direct Energy Formulas

Starting from x = A sin(omega t + phi) and v = omega A cos(omega t + phi), the identity sin^2 + cos^2 = 1 leads directly to compact SHM energy forms. These are powerful because they remove explicit time dependence and let you solve from geometric state alone.

$E_T = \frac{1}{2}m\omega^2A^2, \quad E_p = \frac{1}{2}m\omega^2x^2, \quad E_k = \frac{1}{2}m\omega^2(A^2-x^2)$

These forms are equivalent to spring forms when omega^2 = k/m.

Notice how each formula encodes physical boundaries: at x = 0, kinetic is maximum; at |x| = A, kinetic is zero; and total energy never depends on time in the ideal model. That boundary-check habit is a fast way to catch algebra slips before final answers.

Velocity as a Function of Displacement

$v = \pm \omega\sqrt{A^2-x^2}$

The sign is positive or negative depending on travel direction through the same displacement point.

This formula is especially useful when time is absent from a question. If you know displacement and model parameters, you can compute speed immediately. If direction is required, combine with motion context or displacement-time slope sign to choose plus or minus.

Recovering Unknowns from Maximum Values

$v_{\max}=\omega A, \quad a_{\max}=\omega^2A, \quad \omega=\frac{a_{\max}}{v_{\max}}, \quad A=\frac{v_{\max}^2}{a_{\max}}$

These transformations turn mixed 'max speed/max acceleration' data into period and amplitude quickly.

This data-compression pattern appears frequently in structured-response tasks. Once omega is found, period follows from T = 2pi/omega. Once amplitude is found, any other state quantity can be computed. Treat vmax and amax as a compact signature of one oscillator.

Check Yourself

If your computed displacement ever exceeds amplitude in magnitude, a sign, unit, or substitution error has occurred. SHM always enforces |x| <= A.

Simulation: Quantitative Energy and Speed-Displacement Checks

Stress-test the formulas E_k(x), E_p(x), and v(x) by moving the oscillator through the cycle and verifying consistency at turning points and equilibrium.

Amplitude14.00 cm

Mass0.70 kg

Spring constant42.00 N/m

Time position58.00 % of 2T

Initial phase35.00 deg

Phase offset (2nd oscillator)0.00e+0 deg

Pendulum length1.50 m

Gravity9.81 m/s^2

Period (spring)

0.811 s

Frequency

1.233 Hz

Angular frequency

7.746 rad/s

Period (pendulum, small angle)

2.457 s

Total energy

0.4116 J

Potential energy

0.4108 J (99.8%)

Kinetic energy

8.47e-4 J (0.2%)

In ideal SHM, E_T remains constant while energy shifts between E_p and E_k. At equilibrium (x = 0) kinetic energy is maximum; at turning points (|x| = A) potential energy is maximum.

Test Yourself

An oscillator has vmax = 3.2 m/s and amax = 25 m/s^2. Enter the period in seconds.

Hint: First compute omega = amax/vmax, then T = 2pi/omega.

Your numerical answer (s)

12.3 Energy in Simple Harmonic Motion

Chapter 12 Wrap-Up