In an elevator a system is arranged as shown in figure. Initially elevator is at rest and the system is in equilibrium with middle spring unstretched. When the elevator accelerated upwards, it was found that for the new equilibrium position (with respect to lift), the further extension i the top spring is 1.5 times that of the further compression in the bottom spring, irrespective of the value of acceleration
(a) Find the value of `(m_1)/(m_2)` in terms of spring constants for this happen.
(b) if `k_1=k_2=k_3=500(N)/(m)` and `m_1=2` kg and acceleration of the elevator is `2.5(m)/(s^2)`, find the tension in the middle spring in the final equilibrium with respect to lift.

Explanation

Let the initial extension in top string be `x_1` and the compression in the botton string be `x_3` respectively.
For equilibrium of `m_1`
`k_1x_1o=m_1g` ..(1)
`For equilibrium of `m_2`
`k_3x_3=m_2g` ..(2)
Let the elevator starts acceleration with an acceleration a in the upward direction. Nowin the position of final equilibrium, let further compression in the bottom string be y.
Further extension i the top string`=1.5y`
`and compression in the middle spring`=0.5y` IN the frame of lift
(a) `m_1(g+a)=k_1(x_1+1.5y)+k_2(0.5y)`
`m_1g=k_1x_1`
`m_1a=y((3)/(2)k_1+(1)/(2)k_2)` ..(3)
`m_2(g+a)=k_3(x_3+y)-k_2(0.5y)`
`m_2g=k_3x_3`
`m_2a=y(k_3-(k_2)/(2))` ..(4)
`(3),(4) we get
`(m_1)/(m_2)=(3k_1+k_2)/(2k_3-k_2)`
(b) `k_1=k_2=k_3=500(N)/(m)`
`m_1=2kg`
`a=2.5(m)/(s)`
using (3)
`y=(2xx2.5)/(2xx500)=(1)/(200)m=0.5cm`
Tension in the middle spring`=k(0.5y)`
`=500xx(1)/(2)xx(1)/(2)xx10^-2`
`T=1.25N`