1 hm = ___ m

Text Solution

Verified by Experts

The correct Answer is:
`x=(mkx)/(M+m)``=
-mg-(mkx)/(M+N)`
maximum amplitude
`=(g(M+m))/k`
`a`. From the fre bdoy diagram
:. `R+momegay^2x-mg=0` ……….
Resultant force `momega^2x=mg-R`
`rarr momega^2x=m(k/(M+m))x`
`x=(mkx)/(M+m)`
`omega=sqrt({k/(M+m)})`
[for spring mas system]
`b`. `R=mg-momega^2x`
`=mg-m(mk)/(M+N)x`
`=-mg-(mkx)/(M+N)`
For LR to be smallest `momega^2x` should be max, i.e. x is maximum.
The particle should be at the highest point.
`c`. We have `R=mg-momega^2x`
The two blocks may oscillate together in such a way that R is greater than 0.
At limiting condition R=0
`mg=momega^2x`
`=x=(mg)/(momega^2)=(mg.(M+m))/(mk)`
So the maximum amplitude
`=(g(M+m))/k`
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