A spring executes SHM with mass of 10 kg attached to it. The force constant of spring is 10 N/m. If at any instant its velocity is 40 cm/sec, the displacement will be (where amplitude is 0.5 m )

Options

A. 0.3 m
B. 0.9 m
C. 0.03 m
D. 0.09 m

Answer

A

Explanation

For the spring, `T=2pi sqrt((m)/(K))`
`therefore omega =(2pi)/(T)= sqrt((K)/(m))= sqrt((10)/(10))=1 rad//s`
In S.H.M. `v=omega sqrt(A^(2)-x^(2))`
`therefore v^(2)=omega^(2) (A^(2)-x^(2))=omega^(2)A^(2)-omega^(2)x^(2)`
`therefore (4)/(10)xx(4)/(10)=1 ((1)/(2))^(2)=1xx x^(2)`
`therefore x^(2)=(1)/(4)-(16)/(100)=(9)/(100) therefore x=(3)/(10)=0.3 m`