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

Answer

B

Explanation

`omega=sqrt(k/m)=sqrt((10)/(10))=1`
Since, `v=omega sqrt(a^(2)-y^(2))`
`implies y^(2)=A^(2)-(v^(2))/(omega^(2))=(0.5)^(2)-((0.4)^(2))/(1^(2))`
`implies y^(2)=0.09implies y=0.3m`.