A block of mass one kg is fastened to a spring with a spring constant `50Nm^(-1)`. The block is pulled to a distance `x=10cm` from its equilibrium position at `x=0` on a frictionless surface from rest at `t=0. ` Write the expression for its x(t) and v(t).

Options

A. 5 "cos" (5t)
B. 5 "sin" (5t)
C. 0.5 "cos" (5t)
D. `2xx10^(-2) "cos" (10t)`

Answer

C

Explanation

`omega=sqrt((K)/(m)) " for a spring " [T=2pi sqrt((m)/(K))]`
`therefore omega =sqrt((50)/(2))=5 rad//s`
The block performs a S.H.M. of amplitude A=10 cm =0.1 m, and `omega =5 rad//s`. As it starts from the mean position, its displacemetn equation is `x=A "sin" omega t`
`therefore v=(dx)/(dt)=A omega "cos" (omega t)= 0.1xx5xx "cos" (5t)` ltbr. `therefore v=0.5 "cos" (5t) m//s`