`(1cm)/(1am)` =

Answer

B

Explanation

Let us slove the problem by energy method, iniltial extension of the spring in the mean position `delta=(mg)/k`
During oscilating at ny positin x below the equlibrium positon let tehh velocilty of m be v and angular velocity of the puley be omega. If r is the radius of the pulley, then
`v=romega`
At any instant total energy
=constant (For SHM)
`:. 1/2 mv^2+1/2Iomega^2+1/2k[(x+delta)^2-delta^2]-mgx`
=constant
`rarr(1/2mv^2+(1/2)Iomega^2+(1/2)kx^2+kxd-mgx` ltbrgeconstant
`rarr (1/2)mv^2+(1/2)1(v^2+r^2)+(1/2)kx^2`
=constant `(delta=mg/k)`
Taking derivative of both sides with respect of t
`mv.(dv)/(dt)1/r^2v.(dv)/(dt)+kx(dx)/(dt)=0`
`rarr a((m+1)/r^2)=-kx`
`(:.=(dx)/(dt)and a =(dv)/(dt))`
`rarr a/r=k/((m+1)/r^2)=omega^2`
`rarr T=2pisqrt((m+1/r^2)/k)`