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

The block executes $SHM$, its angular frequency, as given

$\omega =\sqrt{\frac{k}{m}}$

$=\sqrt{\frac{50\, N m ^{-1}}{1 \,kg }}$

$=7.07 \,rad\, s ^{-1}$

Its displacement at any time t is then given by,

$x(t)=0.1\, \cos (7.07\, t)$

Therefore, when the particle is $5\; cm$ away from the mean position, we have

$0.05=0.1 \,\cos\, (7.07\, t)$

$\cos (7.07\, t)=0.5$ and hence

$\sin (7.07 \,t)=\frac{\sqrt{3}}{2}=0.866$

Then, the velocity of the block at $x=5 \,cm$ is

$=0.1 \times 7.07 \times 0.866 \,m s ^{-1}$

$=0.61\, m s ^{-1}$

Hence the $K.E.$ of the block.

$=\frac{1}{2} m v^{2}$

$=1 / 2\left[1\, kg \times\left(0.6123 \,m s ^{-1}\right)^{2}\right]$

$=0.19 \,J$

The $P.E$. of the block,

$=\frac{1}{2} k x^{2}$

$=1 / 2\left(50 \,N m ^{-1} \times 0.05\, m \times 0.05\, m \right)$

$=0.0625 \,J$

The total energy of the block at $x=5\, cm$

$= K.E. + P.E.$

$=0.25 \;J$

we also know that at maximum displacement. $K.E.$ is zero and hence the total energy of the system is equal to the $P.E.$ Therefore, the total energy of the system.

$=1 / 2\left(50 \,N m ^{-1} \times 0.1 \,m \times 0.1 \,m \right)$

$=0.25 J$

which is same as the sum of the two energies at a displacement of $5\; cm$. This is in conformity with the principle of conservation of energy