effective mass of spring

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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}

Joined:10 Apr 2007Posts:1777Thus 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)