Q.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.

[!TLDR] $E=\frac12mv_{max}^2$, so $v_{max}=\sqrt{2E/m}=\sqrt{2(40)/2}=6.324$ m/s (the printed 'cm/s' in the source answer key appears to be a units slip; the …

The speed at the centre (mean position) of the path is the maximum speed $v_{max}$, and the total energy is entirely kinetic there: $E=\frac12mv_{max}^2$ (Eq. 5.24). Given $E=40$ J and $m=2$ kg, $v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(40)}{2}}=\sqrt{40}\approx6.324$ m/s. Note: the source textbook's printed answer states '6.324 cm/s', but with $E=40$ J (SI units, joules) and $m=2$ kg (SI units, kilograms), the computation correctly and unavoidably yields SI units for velocity, i.e. metres per second, not centimetres per second -- this a …

Use $E=\frac12mv_{max}^2$ at the mean position (where all the energy is kinetic) …

Copying the source answer key's unit ('cm/s') without checking dimensional consistency against the SI inputs (jou …