Mass–Spring Systems

On a frictionless track the only horizontal force on the block is the spring's, F = −kx, with x measured from the natural length. Newton's second law gives m d²x/dt² = −kx, which is the SHM equation with ω² = k/m. Nothing else needs solving: for k = 80 N m⁻¹ and m = 0.50 kg, ω = √160 = 12.65 rad s⁻¹, T = 2π/ω = 0.497 s and f = 2.01 Hz. Read the dependence before the numbers. The period goes as √m and as 1/√k, so quadrupling the mass to 2.0 kg only doubles T, to 0.99 s, and halving the period needs a spring four times as stiff. Amplitude does not appear at all: pull the block back 2 cm or 10 cm and it returns on the same schedule, because a larger swing meets a proportionally larger force. Every other system in this topic is solved by reducing it to this one equation with the right k and the right origin.

Now suspend the same spring and let x be its stretch, positive downward. Two forces act, so m d²x/dt² = mg − kx. Set the acceleration to zero and the block rests at Δ = mg/k; for the 0.50 kg block, Δ = 0.50 × 9.81/80 = 0.0613 m. Measure from there instead by writing x = Δ + y. Then mg − k(Δ + y) = (mg − kΔ) − ky = −ky, since the bracket is zero by the definition of Δ. The equation of motion is m d²y/dt² = −ky, the horizontal equation, with the same ω = 12.65 rad s⁻¹ and T = 0.497 s. The reason can be stated without algebra. What sets ω is how fast the net force changes with position, dF/dx = −k. A constant force leaves that slope alone and only slides the point where F = 0 along the axis. The same argument covers a spring on a frictionless incline, where the shift is mg sin θ/k and the period is again unchanged.

Because Δ = mg/k, the ratio m/k that fixes the period equals Δ/g, and so T = 2π√(Δ/g). A ruler therefore predicts the period: a spring that sags 6.13 cm under its load bobs with T = 2π√(0.0613/9.81) = 0.497 s, whatever m and k separately are. The formula has the form of a simple pendulum of length Δ, which makes a handy anchor: a 25 cm sag means a period of about 1 s. It is also the practical route to k. Static method: hang a known m, read Δ, and k = mg/Δ. Dynamic method: time N oscillations and use k = 4π²m/T². If the dynamic value comes out a few per cent lower, you are usually seeing the spring's own mass, which moves too and adds roughly a third of itself to the inertia, while it has no effect on the extra stretch produced by the load.

With x measured from the natural length the potential energy has two parts, U = ½kx² − mgx. Substitute x = Δ + y and use kΔ = mg: U = ½ky² − ½kΔ². The second term is a constant, so measured from equilibrium the hanging oscillator has the single potential ½ky², and gravity need not be mentioned again. Either bookkeeping is correct; mixing them is not. Release the 0.50 kg block from rest at the natural length: it starts a distance Δ above equilibrium, so A = Δ = 6.13 cm, the lowest point is 2Δ = 12.3 cm below the start, and the top speed is vₘₐₓ = ωA = 12.65 × 0.06131 = 0.776 m s⁻¹. The long way agrees: falling Δ releases mgΔ = 0.30074 J, the spring stores ½kΔ² = 0.15037 J, exactly half of it, and the remaining 0.15037 J is ½mv², so v = √(2 × 0.15037/0.50) = 0.776 m s⁻¹ again. Notice that the greatest dynamic stretch, 2Δ, is twice the static one.

Replace a set of springs by the one spring that would pull the mass back just as hard at every displacement; its stiffness is k_eff. If the springs share the displacement, each pulls back with its own kx and the forces add: k_eff = k₁ + k₂. If they are joined end to end, the tension F is common and the extensions add, x = F/k₁ + F/k₂, so 1/k_eff = 1/k₁ + 1/k₂. For 80 and 120 N m⁻¹ that is 200 N m⁻¹ side by side and 48 N m⁻¹ in a chain; with m = 0.50 kg the periods are 0.314 s and 0.641 s. Classify by that test, not by the drawing. A block held between two walls by a spring on each side looks like a chain, but move it x to the right and the left spring pulls left with k₁x while the right spring pushes left with k₂x: shared displacement, so k₁ + k₂. Springs that start out stretched change their forces by the same k₁x and k₂x, so the result stands. The series rule also explains why cutting a uniform spring in half gives two springs of stiffness 2k, since two such halves in series must rebuild the original k.

Three assumptions carry everything above. The spring is massless: a real one of mass mₛ moves with a velocity that rises from zero at the fixed end to the block's speed at the other, and integrating its kinetic energy along the length adds mₛ/3 to the inertia. A 60 g spring under the 0.50 kg block stretches T from 0.497 s to 0.507 s, a 2% effect you can time. The spring pulls and pushes: an elastic cord cannot push, and neither can a spring whose coils already touch, so a vertical oscillation is SHM only while A ≤ Δ. Beyond that the force law fails near the top (a cord goes slack and the block is briefly in free fall), and the period starts to depend on amplitude. The spring is linear: k is the slope of the force–extension line only inside the Hooke's-law range. If a heavy load takes the static stretch past it, the small-oscillation frequency is set by the local slope dF/dx at the equilibrium point, not by the spring's nominal k.