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 = $\sqrt{\frac{2E}{m}}=\sqrt{\frac{2\times 40}{2}}$ = 6.324 m/s