A system is shown in the figure. The time period for small oscillations of the two blocks will be

Options

A. `2pisqrt((3m)/(k))`
B. `2pisqrt((3m)/(2k))`
C. `2pisqrt((3m)/(4k))`
D. `2pisqrt((3m)/(8k))`

Answer

C

Explanation

Both the spring are in series
`:. K_(eq) = (K(2K))/(K + 2K) = (2K)/(3)`
Time period `T = 2pisqrt((mu)/(K_(eq)))`
where `mu = (m_(1)m_(2))/(m_(1) + m_(2))` Here `mu= m/2`
`:. T = 2pisqrt((m)/(2).(3)/(2K)) = 2pi sqrt((mu)/(K_(eq)))`
OR

Total extensin `= 2x`
By energy conservation
`E = 1/2 K_(eq) (2x)^(2) + (1)/(2)mv^(2) + 1/2 mv^(2)`
`E = 1/2 (2k)/3 4x^(2) + 1/2 mv^(2) + 1/2 mv^(2) = 4/3 kx^(2) + mv^(2)`
`(dE)/(dt) = 4/3k(2x) (dx)/(dt) + m(2v)(dv)/(dt)`
there is no loss of energy
`(dE)/(dt) = 0 rArr (8)/(3)kxv + 2mva = 0 rArr (8kxv)/(3) = -2mva`
`a = - (4kx)/(3m) rArr -omega^(2)x = - (4kkx)/(3m) rArr omega =sqrt((4k)/(3m))`
`T = (2pi)/(omega) rArr 2pisqrt((3m)/(4k))`