In the above spring block arrangements, the time period is asked for $K_1=2K$ left spring and $K_2=K$ right spring.
Now at equilibrium, $$T=K_2x=Kx=mg$$ $$2T=K_1y=2Ky=2mg \implies x=y$$ for movable pulley.
How does conservation of mechanical energy hold in spring block systems?
For slow release(no KE change), loss in PE of block is stored in springs.
Since block moves by $2y+x=3x$ down
2y because of movable pulley and x because of right spring
Then
$$mg(2y+x)=\frac{1}{2}K_1y^2+\frac{1}{2}K_2x^2$$
Using my initial force balance equation,
$$mg(3x)=mg(x)+\frac{mg(x)}{2}$$
Therefore energy is not conserved. What happened to the rest of it ? What did I do wrong?
