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

Options

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

Answer

B

Explanation

Time period of spring is given by
`T=2pisqrt(((m)/(k)))`
`impliesomega=(2pi)/(T)=sqrt(((k)/(m)))`
`thereforeomega=sqrt(((10)/(10)))=1`
Velocity of mass at displacement v
`v=omegasqrt(a^(2)-y^(2))`
`impliesy^(2)=a^(2)-(v^(2))/(omega^(2))=(50)^(2)-((40)^(2))/(1^(2))`
`=(50)^(2)-(40)^(2)`
`=(50+40)(50-40)=900`
`thereforey=30cm=0.3m`