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 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.

Explanation

The block executes SHM, its angular frequency, as given by Eq. (14.14b), is
`omega=sqrt(k/m)`
`=sqrt((50Nm^(-1))/(1kg))`
`=7.07rads^(-1)`
Its displacement at any time t is then given by,
x(t) = 0.1 cos (7.07t)
Therefore, when the particle is 5 cm away from the mean position, we have
0.05 = 0.1 cos (7.07t)
Or cos (7.07t) = 0.5 and hence
`sin(7.07t)=(sqrt3)/2=0.866`
Then, the velocity of the block at x = 5 cm is
`=0.1xx7.07xx0.866ms^(-1)`
`=0.61ms^(-1)`
Hence the K.E. of the block,
`=1/2mv^(2)`
`=1/2[1kgxx(0.6123ms^(-1))^(2)]`
`=0.19J`
The P.E. of the block,
`=1/2kx^(2)`
`=1/2(50Nm^(-1)xx0.05mxx0.05m)`
`=0.0625J`
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(50Nm^(-1)xx0.1mxx0.1m)`
=0.25J
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.