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A bar of mass m=0.50kg lying on a horizontal plane with a friction coefficient k=0.10is attached to the wall by means ofa horizontal non- deformed spring. The stiffness of the spring is equal to x=2.45 N//cm, its mass is negligibl . The bar was displaced so that the spring was stretched by x_(0)=3.0 cm, and then released. Find : (a) the period of oscillation of the bar , (b) the total number os osciallations that the bar performs untial it stops completely. |
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Answer» Solution :We shall denote the stiffness constant by `k. ` Suppose the spring is stretched by `x_(0)`. The bar in then subject to two horizontal forces `(1)` restoring force `- kx` and `(2)` friction `kmg` opposing motion. If `x_(0)gt (kgm)/(k )= Delta` the bar will come back. `(` If `x_(0) le Delta`, the bar will stay put . `)` The equation of the bar when it is moving to the left is `m ddot ( x) = - k x + k mg` This equation has the solution `x= Delta + ( x_(0)- Delta) cos sqrt((k)/(m))t` where we have used `x=x_(0), DOT(x)=0` at `t=0` . This solution is onely valid till the bar comes to rest. This happens at `t_(1)=pi//sqrt((k)/(m))` and at that time `x=x_(1)=2 Delta-x_(0)`. if` x_(0)gt 2 Delta` the TENDENCY of the rod will now be to move to the right. `(` if `Deltalt x_(0)lt 2 Delta`the rod will stay put now `)` Now the equation for rightward motion becomes `m ddot(x)=-k x - kmg` `(` the friction force has reversed `)`. We notice that the rod will move to the right only if `k ( x_(0)-2 Delta)gtk mg` `i.e.x_(0)gt3 Delta` In this case the solution is `x=-Delta+( x_(0)-3Delta)cos sqrt((k)/(m))t` Since `x=2 Delta- x_(0) `and `dot(x) =0` at `t=t_(1)=pi //sqrt((k)/(m))`. The rod will next come to rest at `t=t_(2)=2pi //sqrt((k)/(m))` and at that instant `x=x_(2)=x_(0)-4 Delta`. However the rod will stay put unless `x_(0)gt 5 Delta` Thus `(a)` time PERIOD of one full oscillation `=2pi //sqrt((k)/(m)).` `(b)` There is no oscillationif `0 lt x_(0) lt Delta` One half oscillation if `Delta lt x_(0) lt 3 Delta` 2 half oscillation if `3Delta lt x_(0) lt 5 Delta` etc We can say that the NUMBER of full oscillation is one half of the INTEGER`n` where `n=[(x_(0)-Delta)/(2 Delta)]` where`[x]=` smalles non- negative integer greater than `x`. 
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