Oscillations · Lesson 5 of 8
Energy in Simple Harmonic Motion
“Derive the energy relations for SHM, interpret the energy graphs and use conservation to solve an oscillator without finding time.”
• Derive kinetic and potential energies as functions of position and phase. • Show that total mechanical energy is constant for the ideal conservative oscillator. • Explain the choice of potential-energy zero and the T/2 periods of K and U. • Use energy ratios to find displacement, speed or amplitude. • Calculate the source spring-energy example with consistent signs and rounding.
Where does the motion’s energy go at a turning point?
An oscillator is momentarily at rest at either extreme, but its energy has not disappeared. In a spring system, the displaced spring stores potential energy. As the block moves toward equilibrium the spring does work, converting this stored energy into kinetic energy. The block then moves beyond equilibrium and transfers kinetic energy back into spring energy. If resistance is absent and all relevant forces are conservative, the total mechanical energy stays fixed while its division changes.
Kinetic energy follows from the velocity law
Start with K = mv²/2, where m is the oscillating mass and v its signed velocity. Substituting v = −ωA sin(ωt + φ) removes the minus sign because velocity is squared. The relation mω² = k then converts the expression into a stiffness form. K cannot be negative. Its maximum is kA²/2 at equilibrium, and it vanishes at both extremes.
Potential energy comes from the conservative restoring force
A conservative force has a potential energy satisfying F = −dU/dx. For F = −kx, this requires dU/dx = kx. Accumulating the changes from equilibrium to position x gives U(x) − U(0) = ∫₀ˣ kξ dξ. The symbol ξ is just an integration variable running through intermediate displacements. Integration adds the small areas under the force-magnitude-versus-displacement line; that area is a triangle of base |x| and height k|x|. The result is kx²/2 on either side of equilibrium.
Adding a constant C to U changes neither dU/dx nor the force. With U(0) = 0 the potential energy is nonnegative, and the total energy will be positive for nonzero amplitude. A different zero can make some numerical potential energies negative without changing the motion. What matters mechanically is the change in energy and the consistent sum K + U. A vertical spring must include gravitational potential as well as spring potential; about the static equilibrium their combined varying part is again quadratic in the equilibrium displacement.
The sum is constant even though the parts vary
Add the two time-dependent energies. They contain the same factor kA²/2 multiplying cos²θ and sin²θ. Since their sum is one, all phase and time dependence cancels. At the extreme all the variable mechanical energy is potential; at the centre it is kinetic. This gives a useful alternative derivation: the total equals the potential energy at the release extreme, then subtract the potential at any desired x to obtain K without first finding t.
On the time graph U is greatest at 0, T/2 and T, while K is greatest at T/4 and 3T/4. Both have period T/2 because squaring a sine or cosine gives the same value when phase advances by π. Thus energy changes repeat twice during a displacement cycle. On the position graph U rises as an upward parabola, K falls as an inverted parabola within the physically allowed interval, and E is a horizontal line. The curves meet at two positions. A point outside ±A would require negative K from this chosen E and is inaccessible.
Divide the position expressions by E: U/E = x²/A² and K/E = 1 − x²/A². Equal energies require x²/A² = 1/2, so x = ±A/√2, not ±A/2. At x = A/2, U is only one quarter of the total and K is three quarters. Neither energy tells the sign of the velocity: the same energy division occurs while moving inward or outward at that position.
Problem
A 1 kg block on a spring of stiffness 50 N m⁻¹ is released from rest at x = 0.10 m at t = 0. Find K, U and E when x = +0.05 m. Also find the speed and the first-passage velocity.
- 1.The release displacement is the amplitude A = 0.10 m. The angular frequency is √(50/1) = 7.07107 rad s⁻¹, so x(t) = 0.10 cos(7.07107t) m.
- 2.At release K = 0, so E = kA²/2 = 50 × 0.10²/2 = 0.250 J.
- 3.At x = +0.05 m, U = kx²/2 = 50 × 0.05²/2 = 0.0625 J.
- 4.Conservation gives K = E − U = 0.250 − 0.0625 = 0.1875 J. Rounded to two significant figures this is 0.19 J, but use the unrounded 0.1875 J when checking the sum.
- 5.The speed is √(2K/m) = √0.375 = 0.612372 m s⁻¹. On the first trip from +0.10 m toward equilibrium the block is moving left, so v = −0.612372 m s⁻¹. On a later outward crossing at the same positive x it has the opposite velocity.
- 6.As a phase check, cos θ = 1/2 gives θ = π/3 on the first inward trip. Then sin θ = √3/2 and the signed velocity law gives the same negative velocity. This separates speed from direction.
Problem
An ideal oscillator has k = 80 N m⁻¹ and total energy 0.40 J with U(0) = 0. Find its amplitude and its energy division at x = −0.05 m.
- 1.E = kA²/2 gives A = √(2 × 0.40/80) = √0.01 = 0.10 m.
- 2.U = 80 × (−0.05)²/2 = 0.10 J. The sign of x disappears because the spring stores the same energy for equal extension and compression in this ideal model.
- 3.K = 0.40 − 0.10 = 0.30 J. To convert this K into speed a mass is also needed; to determine a full time equation the phase is still needed.
- 4.Changing E with fixed k and m changes A, but not the ideal period 2π√(m/k). This differs from gravitational orbits, whose period changes when their size and energy change.
The constant total here is the mechanical energy of the complete ideal oscillator. With friction, some is transferred to thermal energy, so K plus spring potential decreases even though energy of a larger closed system is conserved. With a driving force, energy can enter the oscillator. Never claim constant oscillator mechanical energy after excluding an external agency that does work.
Quick Quiz
Quiz
With U(0) = 0, where is K = U?
Each energy must equal E/2; U/E = x²/A² then gives |x| = A/√2.
If amplitude doubles for fixed k and m, what happens to E and T?
E ∝ A², while the ideal period depends only on m/k.
What is the period of kinetic energy when displacement period is T?
Squaring the sinusoidal velocity removes the sign reversal after half a displacement cycle.
At x = A/2, what fraction of E is kinetic?
U/E = 1/4 and K/E = 1 − 1/4 = 3/4.
Adding a constant to potential energy does what to the motion?
The derivative of a constant is zero, so F = −dU/dx is unchanged.
Practice Problems
- Derive K(t), U(t) and E for x = A cos(ωt + φ). Explain the use of k = mω² and the trigonometric identity that makes the total independent of time.
- For the 1 kg, 50 N m⁻¹ spring released from +0.10 m, calculate K, U, E and both possible signed velocities at x = +0.05 m. Identify the first-passage velocity.
- An oscillator has amplitude 8 cm. At what displacements are K and U equal? At what displacement magnitudes is K three times U? Use U(0) = 0.
- A mass of 0.20 kg on a spring of stiffness 50 N m⁻¹ has total energy 0.16 J. Find A and vmax. Then find speed at x = A/2.
- Sketch K(t), U(t) and E(t) over one full period for release from a positive extreme. Sketch the corresponding energy-versus-position curves and label their intersections.
- Replace the convention U(0) = 0 by U(0) = −1 J for an oscillator whose varying mechanical energy is 0.25 J. Explain what changes numerically and why the period and forces stay the same.
- Explain why a damped spring can lose mechanical energy without violating energy conservation. State which additional energy store must be included.
Key Takeaways
• K and U exchange continuously while their sum stays constant in the conservative model. • With U(0) = 0, U = kx²/2, K = k(A² − x²)/2 and E = kA²/2. • Each energy part has period T/2, while displacement and velocity have period T. • Energy determines speed magnitude or amplitude but does not determine the direction or phase alone. • A potential-energy zero is a convention; energy differences and forces are the physical quantities.