Short on time? This page is the recap for the calculus-based course (separate from Physics 1). For the proofs, graphs and worked examples, use the full study guide.
Recap
- Take the object–spring system, so the spring’s energy is inside it.
- E = K + U is constant: dE/dt = v_x(m dv_x/dt + kx) = 0 by Newton’s second law.
- At a turning point, E is all potential: E = ½kA². At equilibrium, E is all kinetic: E = ½mv_max².
- K is a maximum when U is a minimum, and the other way round. The minimum K is zero.
- Changing the amplitude changes the total energy; it does not change the period.
Key relationships
| Relationship | Units | Notes |
|---|---|---|
| E = K + U | J | constant with no friction or external work |
| U = ½kx² | J | x from equilibrium |
| E = ½kA² = ½mv_max² | J | gives v_max = A√(k/m) = Aω |
| K = ½k(A² − x²) | J | speed at any position |
| K = U at x = ±A/√2 | m | not at ±A/2 |
| U = ½E[1 + cos 2ωt], K = ½E[1 − cos 2ωt] | J | for x = A cos(ωt); each repeats every T/2 |
| A = √(2E/k) = √(x₀² + mv_x0²/k) | m | amplitude from any starting state |
| Vertical spring: U_total = ½ky² + constant | J | y from the hanging equilibrium |
Assumptions
- Ideal spring, no friction or air resistance, so no energy leaves the system.
- x (or y) measured from the equilibrium position.
- Pendulums: small angles, so U ≈ ½(mg/L)s².
Mistakes to avoid
- E ∝ A. It is E ∝ A²: double A, four times the energy.
- K = U at A/2. At A/2, U = E/4 and K = 3E/4.
- K repeats every T. K and U repeat every T/2.
- More mass, more energy. At fixed A and k, E is the same; v_max is smaller.
- System without the spring. Then the block’s energy changes.
- Vertical spring measured from natural length. Measure from the hanging equilibrium.
Quick self-check
- k = 200 N/m, A = 0.050 m. What is E? (½ × 200 × 0.050² = 0.25 J)
- At what fraction of A is U = E/9? (x = A/3)
- A block has v_max = 0.90 m/s. What is its speed at x = A/2? (v = v_max√(1 − ¼) = 0.78 m/s)
Next: practice questions.