Spring of spring constant `k` is attached with a block of mass `m_(1)`, as shown in figure. Another block of mass `m_(2)` is placed angainst `m_(1)`, and both mass masses lie on smooth incline plane.

Find the compression in the spring when lthe system is in equilibrium.

Options

A. `((m_(1) + m_(2))gsintheta)/(2k)`
B. `((m_(1) + m_(2))gsintheta)/(k)`
C. `((m_(1) + m_(2))g)/(k)`
D. `(2(m_(1) + m_(2))gsintheta)/(k)`

Answer

B

Explanation

At equilibrium position : `rarr`
`Kx_(0) = m_(1)g sintheta + m_(2)g sintheta`
`x_(0) = ((m_(1) + m_(2))gsintheta)/(k)`
Block will separate when acceleration of block of mass `m_(1)` will be just equal to `g sin theta`
`rArr a = omega^(2)x = g sintheta`
`rArr (K)/(m_(1) + m_(2)) xx x = g sin theta`
`x = (m_(1) + m_(2))/(K)g sintheta = x_(0)`
So, blocks will separate when blocks are at distance `x = (m_(1) + m_(2))/(K) g sintheta`
From mean position
i.e. spring is at its natural position (amplitude `A = ((2)/(K)(m_(1) + m_(2))g sintheta')`)
Blocks will seoparte at distance `x = x_(0)` from mean position.
`v = sqrt(A^(2) - x_(0)^(2))`
`= sqrt((K)/(m_(1) + m_(2))) sqrt(((2)/(K)(m_(1) + m_(2))g sintheta)^(2) - (((m_(1) + m_(2))/(K))g sintheta)^(2))`
`= sqrt((3)/(k)(m_(1) + m_(2)))gsintheta`