Question

A body of mass `1kg` is suspended from a weightless spring having force constant `600N//m`. Another body of mass `0.5 kg` moving vertically upward hits the suspended body with a velocity of `3.0m//s` and get embedded in it. Find the frequency of oscillations and amplitude of motion.

Answer 1

Here inertia factor `=m+M`
`=1+0.5=1.5kg`
Spring factor `=k=600Nm^(-1)`
`:.` Frequency, `v=(1)/(2pi)sqrt((k)/((m+M)))`
`=(7)/(2xx22)sqrt((600)/(1.5))=3.2 Hz`
Using law of conservation of linear momentum in the collision
`mv=(m+M)V`
or `V=(mv)/((m+M))=(0.5xx3)/(1+0.5)=1ms^(-1)`
Now just after collision, the system will have maximum K.E. `= (1)/(2)(m+M)V^(2)`
Using law of conservation of mechanical energy we have,
`(KE)_(max)=(PE)_(max)=E`
So `(1)/(2)(m+M)V^(2)=(1)/(2)ka^(2)`
or `a=Vsqrt((m+M)/(k))=1sqrt((1.5)/(600))=(1)/(20)m=5cm`