1Gm =__nm

Answer

A, B

Explanation

Let the amplitude of oscillation of m and M be `x_1 and x_2` respectively.
a. From law of conservation of momentum `mx_1=Mx_2` ………1
[Block and mass oscillates in opposite direction bt `xprarr` stretched part]. From equation 1 and equation 2`
`:. x_0=x_1+m/Mx_1`
`=((M+m)/M)x`
`so, x_2=x_0-x_1`
`x_0[1-M/(M+m)]`
`=(mx_0)/(M+m)` respectively
b. At any position, let the velocity be `v_1 and v_2` respectively. Here `v_1 velocity of m'` with respect to M. By energy method,
Total Energy=Constant
`1/2mv_2^2+1/2m(v_1-v_2)^2+1/2k(x_1+x_2)^2` ltbr.gt =constant.......3
`[v_1-v-2=` absolute velocity of mass m as seen from the road]
`Mx_2=mx_1`
`rarr x_1=(M/m)x_2`.........4
`Mv_2=m(v_1-v_2)`
`rarr (v_1-v_2)=(M/m)v_2`..........5
Putting the above values in equation 3 we get
`1/2Mv_2^I2+1/2m M/m^2 v_2^2+1/2kx_2^2(1+M/m)`
`:.M(1+M/m)v_2^2+k(1+M/m)x_2^2=constant`
`rarr Mv_2^2+k(1+M/m)x_2^2=constant`
Taking derivative of both sides,
`M_2v_2(dv_2)/(dt)=k((M+m)/m)2x_2(dx_2)/(dt)=0`
`rarr ma_2+k((M+m)/m)x_2=0`
`[because v_2=(dx_2)/(dt)]`
`a_2/x_2=(-kM+m)/(Mm)omega^2`
`:. omega=(-k(M+m))/(Mm)`
So time period
`T=2pi sqrt((Mm)/(M+m))`