Springs, Hooke's Law & Combining Stiffnesses
- Apply Hooke's law as a force law with correct sign conventions
- Derive the effective stiffness of springs in series and in parallel
- Analyze equilibrium positions for systems involving springs and gravity
Hooke's law is a restoring force
F = −kx says that the force a spring exerts is proportional to its displacement from the natural length and directed opposite to that displacement. The minus sign is the physics: it makes the force restoring, always pushing or pulling the system back toward equilibrium, which is precisely the condition that produces simple harmonic motion later in the course. The spring constant k has units of N/m and measures stiffness — a large k means a stiff spring.
Parallel and series combinations
Two springs in parallel both stretch by the same amount x and their forces add, so F = (k₁ + k₂)x and k_parallel = k₁ + k₂. Two springs in series carry the same force while their extensions add, so x = F/k₁ + F/k₂ and 1/k_series = 1/k₁ + 1/k₂. The combinations behave oppositely to the intuition many students bring: adding a spring in parallel makes the system stiffer, while adding one in series makes it softer than either spring alone.
Hanging mass and the shifted equilibrium
Hang a mass m from a vertical spring and it settles where the spring force balances gravity: kx₀ = mg, so the new equilibrium is x₀ = mg/k below the natural length. The important consequence appears in oscillation problems: if you measure displacement from this new equilibrium rather than from the natural length, the gravitational term cancels out of the equation of motion entirely, and the system obeys exactly the same F = −ky as a horizontal spring. That is why a vertical mass-spring system has the same period as a horizontal one.
Springs with k₁ = 200 N/m and k₂ = 300 N/m are combined. Find the effective stiffness in parallel and in series. Then find how far a 2.0 kg mass hangs below the natural length of the 200 N/m spring alone.
- 1.Parallel: k_eff = k₁ + k₂ = 200 + 300 = 500 N/m.
- 2.Series: k_eff = k₁k₂/(k₁ + k₂) = (200)(300)/500 = 60000/500 = 120 N/m.
- 3.Check the series result: 120 N/m is smaller than either 200 or 300, as it must be.
- 4.Hanging mass: x₀ = mg/k = (2.0)(9.8)/200 = 19.6/200 = 0.098 m.
Two identical springs of constant k are connected end to end in series. The effective spring constant of the combination is —
Series and parallel spring formulas are the reverse of the resistor formulas and the same as the capacitor formulas. If you remember that resistors in series add, note that springs in parallel add, and you will not mix them up.
A mass hangs at rest from a vertical spring. If displacement is measured from this hanging equilibrium position rather than from the spring's natural length, the equation of motion —
Answer the 2 checkpoints as you read.
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