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Home/Classical Mechanics/Slinky Drop (Springs)

Slinky Drop (Springs)

The Slinky Drop simulator visualizes a classic physics demonstration where a suspended Slinky is released from its top. It models the Slinky as a one-dimensional chain of point masses connected by ideal, massless springs. Each spring obeys Hooke's Law, F = -kΔx, where k is the spring constant and Δx is the displacement from equilibrium. When the top anchor point is released, gravity acts on every mass. However, the bottom of the chain does not begin to fall immediately. This counterintuitive lag occurs because the information that the top is no longer supported must propagate down the chain as a longitudinal stress wave. The wave speed depends on the spring constant and the mass of each segment. The simulator solves Newton's second law, F_net = ma, for each mass in the chain at each time step, calculating the net force from the springs above and below and from gravity. Key simplifications include neglecting air resistance, damping, and any transverse motion, focusing purely on the 1D vertical dynamics. By interacting with the model, students can explore the concepts of wave propagation in a dispersive medium, the finite speed of information transfer, and how a system's internal forces respond to a changing boundary condition. It provides a clear, visual connection between microscopic interactions (spring forces) and macroscopic behavior (the delayed collapse).

Who it's for: High school and introductory college physics students studying Newtonian mechanics, waves, and oscillations.

Key terms

  • Hooke's Law
  • Stress Wave
  • Newton's Second Law
  • Wave Propagation
  • Longitudinal Wave
  • Spring Constant
  • Dispersive Medium
  • Equilibrium Position

Chain

48

Masses and ideal springs under gravity: the famous “bottom floats” is a stress-wave effect, not anti-gravity.

Measured values

Coils18

How it works

A minimal discrete slinky: enough physics to communicate delayed motion at the lowest turns.

Frequently asked questions

Why doesn't the bottom of the Slinky fall immediately when I drop it?
The bottom mass is initially in equilibrium, held up by tension from the spring above it. When the top is released, the tension in the top spring vanishes, but this change must travel down as a wave. Until that wave of reduced tension reaches the bottom, the spring directly above the bottom mass still exerts its original upward force, so the bottom mass remains momentarily suspended. This demonstrates that forces and information do not act instantaneously through a material.
Is this a realistic model of a real Slinky?
It captures the essential physics but makes simplifications. A real Slinky has mass distributed along its coils, experiences air resistance, and can twist and bend. Our model simplifies it to point masses and massless springs in one dimension, which isolates and clarifies the core mechanism of wave propagation without complicating factors. The qualitative behavior—the bottom lag—is accurately represented.
What determines how fast the wave travels down the chain?
The wave speed depends on the stiffness (spring constant, k) and the inertia (mass per segment, m) of the chain. For a discrete spring-mass system, the speed is proportional to sqrt(k/m). A stiffer spring (larger k) transmits forces faster, while a heavier mass (larger m) responds more sluggishly, slowing the wave. In a continuous, uniform spring, the speed is sqrt(kL/m_total), where L is the length.
Does gravity affect the wave speed in this simulation?
No, gravity does not affect the speed of the stress wave itself. Gravity provides a constant background force on each mass, setting the initial stretched equilibrium. The wave speed is determined by the elastic properties (k) and inertia (m) of the system. However, gravity is crucial for creating the initial tension and for the overall falling motion after the wave passes.