AI Tutor
When a wrecking ball reaches the top of its swing, it freezes motionless for a split second before roaring back down with terrifying speed.
Where does all that blazing kinetic energy disappear to, and how does nature seamlessly store and return it without losing a single joule in an ideal system?

Mechanical Energy in a Spring-Mass System

In simple harmonic motion, a system continuously converts energy back and forth between kinetic energy (energy of motion, KK) and potential energy (stored energy, UU).
For a mass mm attached to a spring with spring constant kk, the elastic potential energy is U=12kx2U = \frac{1}{2}kx^2 and kinetic energy is K=12mv2K = \frac{1}{2}mv^2.
When no friction or air drag is present, the total mechanical energy (E=K+UE = K + U) remains strictly constant throughout the oscillation.
At the turning points (x=Β±Ax = \pm A), the mass stops instantly (v=0v = 0), so the total energy is purely potential: E=12kA2E = \frac{1}{2}kA^2.
At the equilibrium position (x=0x = 0), the spring is completely unstretched (U=0U = 0), meaning all energy has transformed into maximum kinetic energy: E=12mvmax2E = \frac{1}{2}mv_{\text{max}}^2.
Equating both extreme energy states gives the formula for maximum speed: 12mvmax2=12kA2β€…β€ŠβŸΉβ€…β€Švmax=Akm=Ο‰A\frac{1}{2}mv_{\text{max}}^2 = \frac{1}{2}kA^2 \implies v_{\text{max}} = A\sqrt{\frac{k}{m}} = \omega A, where Ο‰\omega is the angular frequency.
What happens to the speed and energy when the mass is halfway between the center and the maximum stretch point?

Energy as a Function of Position

At any position xx, conservation of energy requires that 12mv2+12kx2=12kA2\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2.