Question

A particle mass m is attached to two identical springs each of spring constant k and natural length a. One spring is attached at its other end to a fixed point A, the other spring to a fixed point B such that A is a height 4a vertically above B. The particle can move in the vertical line AB. If mg/2 12 years ago

Answer 1

I assume that my 2nd spring is still stretched upwards so the free body diagram looks like this:

They're in equilibrium so Newton's second law:
\[k(x+a)=mg+k(a-x) \rightarrow x=\frac{ mg }{ 2k}\]
independent of a.
This is the case when force due to 2nd spring is downwards. And this is possible when
\[k(a-x)>0 \rightarrow x<a\]
To prove that we know \[\frac{ mg }{ 2}<ka\]
and\[\frac{ mg }{ 2 }=kx\]so \[kx<ka \rightarrow x<a\]
In addition (it's not the part of the answer) I'd like to see what's going to happen when x=a
and x>0. In first case this is only true when
\[a=x=\frac{ mg }{ 2k }\] where a is length of spring
if \[x>a\]
\[k(2a+x)+kx=mg \rightarrow x=\frac{ mg }{ 2k }-a\]