A 1.4 kg mass is attached to a spring with a force constant of 56 N/m. If the mass is released with a speed of 0.26 m/s at a distance of 8.6 cm from the equilibrium position of the spring, what is its speed when it is halfway to the equilibrium position?
Let X = distance from the equilibrium position and k = spring constant.
(1/2)kX^2 + (1/2) M V^2 = constant
= 28*(.086)^2 + (0.7)(0.26)^2
= 0.2071 + 0.0473 = 0.2544 joules
Amplitude (maximum deflection) Xmax is where V = 0, so
(1/2)kXmax^2 = 0.2544
Xmax = 0.0953 m = 9.53 cm
Calculate V when X = Xmax/2
3/4 of the total energy will be kinetic at that point.
(M/2)V^2 = 0.75*(0.2544 J)
Solve for V