DART GUN
½kx² becomes ½mv² becomes mgH — one joule in three costumes, and it is shm’s peak speed.
Use the simulation above to change the variables and play through the guided stages. The explanation below describes the default starting values; the simulation updates its explanation as you experiment.
The question
A spring of stiffness k is compressed a distance x and launches a dart of mass m straight up. How high does the dart rise before gravity turns it around?
Spring → motion
The elastic energy stored in the compressed spring, ½kx², is handed entirely to the dart as kinetic energy ½mv² at the instant it leaves the barrel. No energy is lost.
Solve
Equating the two energies gives the launch speed v = x√(k/m). From there the dart is a projectile: it rises until all its kinetic energy has become gravitational, at height H = v²/2g.
The flight
Up it goes, kinetic energy pouring into the gravitational bar and speed bleeding away. At the apex the dart is momentarily still: every joule now sits in height, none in motion.
One energy
The same joule wears three costumes: elastic ½kx², kinetic ½mv², gravitational mgH. And the launch speed is exactly the peak speed of a spring oscillator swinging with amplitude x.