A spring connects two particles of masses `m_(1)` and `m_(2)` A horizontal force `F` acts on `m_(1)` Ignoring friction when the elongation of the spring is `x` then [when the spring has maximum elongation]
.

Options

A. `a_(2)=Kx//m_(2)`
B. `a_(1) =(F -Kx)//m_(1)`
C. `F =m_(1) a_(1) + m_(2) a_(2)`
D. `a_(1) =a_(2) = (F)/((m_(1) + m_(2))`

Answer

a,b,c,d

Explanation

Here, (a) Restoring force on spring `=-Kx`
for mass `m_(2)`
`m_(2) a_(2) - Kx =0`
`m_(2) a_(2) = Kx`
`a_(2) = (Kx)/(m_(2))`
(b) From mass `m_(1)`
`F - Kx = m_(1) a_(1)`
`:. A_(1) = (F -Kx)/(m_(1))`
(c) Adding `F - Kx + Kx = m_(1) a_(1) + m_(2) a_(2)`
`F = m_(1) a_(1) + m_(2) a_(2)`
(d) If `a_(1) = a_(2)`
`F = (m_(1) + mm_(2))a` or `a = (F)/(m_(1) + m_(2)` .