Example

Question:

A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of \(50\,\mathrm{N\,m^{-1}}\). The block is pulled to a distance \(x=10\,\mathrm{cm}\) from its equilibrium position at \(x=0\) on a frictionless surface from rest at \(t=0\). Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.

Solution:

The block executes SHM. Angular frequency: \[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{1}} = 7.07\,\mathrm{rad\,s^{-1}} \] Displacement at any time \(t\): \[ x(t) = 0.1 \cos(7.07 t) \] At \(x = 5\,\mathrm{cm} = 0.05\,\mathrm{m}\):
\(\cos(7.07 t) = 0.5\), so \(\sin(7.07 t) = \sqrt{3}/2 = 0.866\).
Velocity: \[ v = -A\omega \sin(\omega t) = 0.1 \times 7.07 \times 0.866 = 0.61\,\mathrm{m\,s^{-1}} \] Kinetic energy: \[ KE = \frac{1}{2}mv^2 = \frac{1}{2}\times1\times(0.6123)^2 = 0.19\,\mathrm{J} \] Potential energy: \[ PE = \frac{1}{2}k x^2 = \frac{1}{2}\times 50 \times (0.05)^2 = 0.0625\,\mathrm{J} \] Total energy at \(x = 5\,\mathrm{cm}\): \[ E = KE + PE = 0.19 + 0.0625 = 0.25\,\mathrm{J} \] At maximum displacement (\(x=10\,\mathrm{cm}\)), total energy is all potential: \[ E = \frac{1}{2}kA^2 = \frac{1}{2}\times 50 \times (0.1)^2 = 0.25\,\mathrm{J} \] This confirms conservation of energy.