Question

The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. 

Answer 1

Given : m = 2 kg, E = 40 J

The speed of the body while crossing the centre of the path (mean position) is v_max and the total energy is entirely kinetic energy.

  \(\frac{1}{2}\)mv²_max = E

∴  v_max = \(\sqrt{\frac{2E}{m}}= \sqrt{\frac{2×40}{2}}\) = 6.324 m/s