In an ideal mass-spring system (oscillating horizontally on a frictionless surface), the total energy is conserved.
Using the fact that energy is conserved, find an expression for maximum speed in terms of m, k and/or A .
We know that the simplest equation of motion for an oscillating mass-spring system is of sinusoidal form having an initial phase zero.
The speed of the mass is the first derivative of the elongation
`v(t) =dy/dx = A*omega*cos(omega*t) rArr v_("max") =A*omega`
Thus the potential end kinetic energies are
`E_p =-ky^2/2 =k*A^2/2 *sin^2(omega*t)`
`E_k =m*v^2/2 =(m*omega^2)*A^2/2*cos^2(omega*t)`
In the case there are no energy losses we have
`E_p(max) = E_k(max)`
`k*A^2/2 =(m*omega^2)*A^2/2` , which means `omega^2 =k/m`
Thus the maximum speed is
`v_max =A*omega =A*sqrt(k/m)`
Answer: the maximum speed of the oscillator is `v_max =A*sqrt(k/m)`