There are two ideal springs of force constants `K_1` and `K_2` respectively. When both springs are relaxed the separation between free ends is L. Now the particle of mass m attached to free end of left spring is displaced by distance 2L towards left and then released. assuming the surface to be frictionless. `((K_(1))/(K_(2))=4/3)`. (Neglect size of the block)

Suppose mass m hits the right spring and sticks to it. The extension in left spring when mass ‘m’ is in equilibrium position during its motion is :

Options

A. 4L/7
B. 3L/7
C. L/3
D. L/2

Explanation

Time taken by particle to go from
x=0 to x=A/2 is T/12
`:. ` time interval `=T/2+T/12=(7T)/12`
`=7/12 . 2pisqrt(m/(K_(1)))=(7pi)/6sqrt(m/(K_(1)))`
Assume, maximum compression in right spring is x. hence
`1/2 K_(1)(2L)^(2)=1/2 K_(1)(L+x)^(2)+1/2K_(2)x^(2)`
Put `K_(2)=3/4 K_(1)`, we get `x=(6L)/7`
When mass m in equilibrium both spring will be in extended state.

`K_(1)x_(1)=K_(2)x_(2)` and `x_(1)+x_(2) =L`
`x_(1)=(3L)/7`.