A block whose mass is 2 kg is fastened on a spring whose spring constant is 100 `Nm^(-1)` . It is pulled to a distance of 0.1 m from over a frictionless surface and is released at t=0. Calculate the kinetic eneryg of the block when it is 0.05 m away from its mean position.

Explanation

The block executes SHM so, its angular frequency, `omega=sqrt((k)/(m))=sqrt((100Nm^(-1))/(2 kg))=7.07 "rads"^(-1)`
Its displacement at any time t is
`x(t)=a " cos"omegat=0.1 "cos"(7.07t)`
When the particle is 0.05m away from the mean position.
`0.05=0.1 "cos"(7.07t)`
or cos(7.07t)=0.5
or `"sin"(7.07t)=(sqrt(3))/(2)=0.866`
Velocity of the block at x=0.05m is
`v=Aomega"sin"omegat`
`0.1 xx7.07 xx 0.866=0.61 ms^(-1)`
Hence, `KE=(1)/(2)mv^(2)=(1)/(2)xx2(0.61)^(2)=0.37 J`