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Mechanics

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 Joined: 10 Jul 2007 Post: 18
30 Jan 2008 15:52:13 IST
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effective mass of spring
Engineering Entrance , Medical Entrance , NEET , JEE Main , AIIMS , JEE Main & Advanced , Physics , Mechanics

In the derivation of finding the time period of a system inn which the spring also has some mass we take v(velocity at a infinitismal part at a distance x from the point of suspension)=(x*V)/L where V is the velocity of mass attached to the spring at its lowermost point,and L is the normal length of the spring....how to derive this expression of v(velocity at a infinitismal part)......

given derivation for an infinitesimal element of length dx of the spring at a distance of x from the fixed end velocity=(v*x)/l.
E=1/2*mv^2  +  1/2*kx^2   +   int{1/2*m'/l*[(v*x)/l]^2}dx.
as in shm E=constant,dE/dt=0.
take the derivative w.r.t time yull get a=-k/(m+m'/3)*x.
so T=2*pi*sqrt{(m+m'/3)/k}

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Joined: 10 Apr 2007
Posts: 1777
30 Jan 2008 16:52:56 IST
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Throughout the spring the stress remains constant, so the strain must too. All portions of the spring have different extensions and different initial lengths, but as the strain is constant, the ratio of extension to the initial length is a constant. Here the extension of any part of the spring is its displacement and initial length is the element's initial position.
Thus for an infinitesimal displacement of any element located at a position x from the fixed end of the spring,
dx/x = constnt
(dx/dt)/x = constant
v/x = constnt
so,
V/L = v/x
v = V(x/L)

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Joined: 1 Feb 2008
Posts: 1
1 Feb 2008 00:52:53 IST
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You have mentioned that the stress in the spring remains const. but stress=F/A ... but the Force is the Tension in the spring and since the spring is not massless the Tension continues to vary throughout the spring...the stress also varies throughout

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