Question

The spring as shown in figure is kept in a stretched position with extension `x` when the system is relesed. Assuming the horizontal surface to be frictionless, the frequency of oscillation is

Answer 1

Considerign the two blocks plus the spring as a system, there is o exterN/Al resultant force on the system. Hence the centre of mass of the system will remain at rest. tE mean positions of the two simple harmonic motions occur whe the spring becomes unstretched. If the mass m moves towasrds right through a distance x and the mass M moves towards left through a distance X before the spring acquires N/Atural lenght
`x+X=x_0`.............i
x and X will be the amplitudes of the two blocks m and M respectively. As the centre of mass should not change during the motion, we should also have
`mx=MX`............ii ltbrogt From i and ii `x=(Mx_0)/(M+m)and X=(mx_0)/(M+m)`
Hence the left block is `x=(Mx_0)/(M+m)` distance away from its mean position in the beginning of the motion. the force by the spring on this block at this instant is equal to the tension of spring
`T=kx_0`
Now `x=(Mx_0)/(M+m) or , x_0=(M+m)/Mx`
`Thus, T=(k(M+m))/Mx or, a=T/m=(k(M+m))/(Mm)x`
`The angular frequency is therefore `omega=(sqrt(k(M+m))/(Mm))`
and the frequency is `v=omega/(2pi)=1/(2pi)(sqrt((M+m))/(Mm))`