A body of mass 1kg is suspended from a spring of negligible mass. Another body of mass 500g moving vertically upwards hits the suspended body with a velocity of `3 ms^(-1)` and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision is `(10)/(pi) Hz`, the amplitude of the motion and the spring constant are respectively,

Options

A. `5 cm, 300 Nm^(-1)`
B. `10 cm, 300 Nm^(-1)`
C. `10 cm, 600 Nm^(-1)`
D. `5 cm, 600 Nm^(-1)`

Answer

D

Explanation

Given, mass of body, `M = 1 kg`, frequency of oscillation, `f = (10)/(pi) Hz`,
mass of hitting body, `m = 0.5 kg` and speed of body, `v = 3ms^(-1)`
As frequency of oscillation of spring mass system,
`f = (1)/(2pi) sqrt((k)/(M_("body")))`
Since, hitting body get embedded to the initial mass, so mass of the body changed to
`M_("body") = M + m = 1 + 0.5 = 1.5 kg`
Spring constant, `k = (2pi f)^(2)M_("body")`
Now, velocity of body after collision,
`(M + m) v. = m v rArr (1 + 1//2) v. = 1//2 xx 3`
`v. = 1m//s`
The maximum amplitude of the sping-mass system is `v. = A omega rArr A = (v.)/(2pi f) = (1)/(20) = 0.05m = 5m`
Hence, the correct answer is (d).