The block of mass `m_1` shown in figure is fastened to the spring and the block of mass `m_2` is placed asgainst ilt. A. Find the compression of the spring in the equilibrium position. b.The blocks are pushed a further distance `(2/k)(m_1+m_2)gsintheta` against the spring and released. Find the position where the two blocks separate. c. What is the common speed of blocks at the time of separation?

Options

A. `(sqrt((1)/(3k)(m_(1)+m_(2))))gsintheta`
B. `(sqrt((2)/(k)(m_(1)+m_(2))))gsintheta`
C. `(sqrt((3)/(k)(m_(1)+m_(2))))gsintheta`
D. `(sqrt((1)/(k)(m_(1)+m_(2))))gsintheta`

Answer

C

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`