A body of mass 1.0 kg is suspended from a weightless spring having force constant `600Nm^(-1)`. Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of `3.0ms^(-1)` and gets embedded in it. Find the frequency of oscillations and amplitude of motion.

Answer

A, C

Explanation

Frequency of oscillation
`f = (1)/(2pi) sqrt((k)/(m_(1) + m_(2))) = 1/(2pi) sqrt((600)/(1.5)) = 10/pi Hz`
Let maximum amplitude ba A than
`v = omegasqrt(A^(2) - x^(2))`
where `x =` difference in equilibrium position
`= ((m_(1) + m_(2))g)/(k) - (m_(1))g/(k) = 1/120 m`
and `v = (0.5 xx 3)/(1.5) = 1 m//s`
Therefore
`1 = 20 sqrt(A^(2) - (1/120)^(2)) rArr A = (5sqrt(37))/(600) = (5sqrt(37))/(6) cm`