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All the surfaces shown in figure are frictionless. The mass of the car is M, that of the block is mk and the spring has spring constant. Initialy the car and the block are at rest and the spring is stretched through a length x_0 when the system is released. a. Find the amplitude of the simple harmonic motion of the blocks and of the car as seen from the road. b. Find the time periods of the two simple harmonic motions. |
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Answer» 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))` |
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