Explain spring turning points and fastest motion

Follow the first half of the journey

The block begins at +40 cm, released from rest. Its spring pulls left. As it approaches equilibrium, restoring force becomes smaller, yet speed becomes greater. At equilibrium the spring force is momentarily zero and speed is greatest. The block continues left because it already has velocity; no force is required to preserve an instantaneous velocity.

After crossing equilibrium, the spring is compressed and pushes right. The block still moves left, so the force reduces its speed. It reaches −40 cm at rest, then accelerates back toward the right. A turning point is a moment of zero velocity, not a command that holds the block stopped.

Use stiffness 20 N/m. Predict the force at +40, 0 and −40 cm. Then visit those positions in that order and save captures. The values are −8, 0 and +8 N. With mass 1 kg, accelerations are −8, 0 and +8 m/s². Compare those results with the speeds instead of assuming acceleration and speed always increase together.

Balance the energy of the block and spring

Choose the block–spring system. The fixed support does not move, the track is frictionless, and the spring is ideal. Its mechanical energy is E = K + U, the sum of the block’s translational kinetic energy and elastic potential energy. Because height does not change, gravitational potential differences do not enter this exchange.

The release has amplitude A = 0.40 m and zero kinetic energy. At k = 20 N/m, E = ½ × 20 × 0.40² = 1.6 J. This number remains constant during that release. At x = +0.20 m, U = 0.4 J and K = 1.2 J. The displayed gap between the potential curve and total line is that 1.2 J.

Kinetic energy is K = ½mv². Here m is mass in kilograms and v² is squared speed in metres squared per second squared. Recover speed using speed = √(2K/m). At +0.20 m, speed = √2.4 ≈ 1.549 m/s. At equilibrium U = 0, K = 1.6 J and speed = √3.2 ≈ 1.789 m/s. At either turn, U = 1.6 J and K = 0.

Energy conservation alone supplies speed, not the sign of velocity. The first half-cycle in this widget runs from right to left, so interior velocities are negative. On the later return pass the block has the same speed and energies at a given position but positive velocity. The initial condition and which pass is selected resolve the ambiguity.

Separate force, velocity and power

At +20 cm, force and first-pass velocity are both negative: the spring does positive work as the block moves left and kinetic energy increases. At −20 cm, force is positive while velocity is negative: spring work is negative and kinetic energy decreases. The spring still points toward equilibrium on both sides.

At equilibrium the instantaneous force is zero. An instant of zero acceleration does not mean acceleration remains zero afterward. Moving a little farther left makes the spring compressed and its force positive. At the left turning point, velocity is zero but the nonzero positive acceleration starts the return motion.

This distinction also explains why the block cannot remain motionless at a nonzero displacement unless another force is added. Holding it compressed requires an external balancing force. The lab contains no brake, latch or holding controller.

Interpret period without treating position as a clock

The period T is the time to return to the same position and direction of motion after one whole oscillation. For this ideal model, T = 2π√(m/k). With 1 kg and 20 N/m, T ≈ 1.405 s. The first center crossing occurs at T/4 ≈ 0.351 s; the left turn at T/2 ≈ 0.702 s.

The position slider selects a state, not an equal time step. Equal displacement increments take longer near a turn, where speed is low. The lab’s first-arrival time uses the exact oscillator solution for release at +A. You can understand the energy investigation without deriving that solution.

Explore the optional time relationship

Define angular frequency ω = √(k/m), measured in radians per second. The release solution is x(t) = A cos(ωt), where t is seconds since release. On the first half-cycle, 0 ≤ t ≤ π/ω, the inverse gives t = arccos(x/A)/ω. The inverse cosine uses radians. This branch is why the displayed time describes first arrival rather than every later visit. Differentiating the solution gives acceleration −ω²x, consistent with F/m = −kx/m.

Double stiffness to 40 N/m with the same mass and amplitude. The preparation stores twice the energy, center speed increases by √2, and period decreases by √2 to about 0.993 s. This comparison changes initial preparation energy; it is not an energy-conserving stiffness adjustment during one oscillation.

Transfer beyond the controls

A 2 kg block attached to a 50 N/m ideal spring is released from rest 0.30 m from equilibrium. Predict its total energy, speed at equilibrium and acceleration at the positive turning point. These mass and amplitude values differ from the widget; solve them independently.

Check the spring-motion transfer

E = ½ × 50 × 0.30² = 2.25 J. At equilibrium speed = √(2 × 2.25 / 2) = 1.5 m/s. At the positive turn, force is −15 N, acceleration is −7.5 m/s² and speed is zero. The full period is 2π√(2/50) ≈ 1.257 s. None of these results requires treating force as constant through the journey.

A real oscillator may lose mechanical energy through friction or internal deformation. Then successive turning amplitudes can shrink and K + U alone is not constant. Those losses require an expanded accounting model; this undamped lab does not fit measured decay or verify a real spring’s properties.