*A massless spring of free length L and spring constant K is stretched until a desired force F is achieved. A mass M is attached to the stretched spring, and the spring is then allowed to accelerate the mass until the spring length is L2.
Find a closed-form solution for the speed of the mass when the spring length is L2 (where L2>L). Ignore friction and gravity. Use the solution to explore the effect of small changes in L and K (with F, M, and L2 held constant).
*
*
Energy in the system = F^2 / 2K
Length when stretched (L3) = L + F/K
Potential energy at L2 = 0.5K*(L2-L)^2
Kinetic energy @L2 = F^2 / 2K - 0.5K*(L2-L)^2
Velocity@L2 = Sqrt(2/M( (F^2 / 2K) - 0.5K*(L2-L)^2 ))
=Sqrt(2K/M( (F/K)^2 - 0.5(L2-L)^2))
I don’t know what to do with that.
I think it should be v=sqrt( F^2/(km)-(k/m)(L2-L)^2).
Did you mean F^2/(2K) ?
*
*
Could you show your work please?
*
*
Sure. I used a somewhat roundabout method. Let A = F/k be the amplitude of the oscillation. The position will follow simple harmonic motion with equation x = L + A * cos(sqrt(k/m)*t), with x = 0 as the left end of the spring and the mass started at t = 0. So we set x = L2 and so L2-L = A * cos(sqrt(k/m)t). Also, v = x’ = - A * sqrt(k/m) * sin(sqrt(k/m)t). So then we have v^2 = A^2 * (k/m) * sin^2(sqrt(k/m)t) = (k/m) * (A^2- A^2 * cos^2(sqrt(k/m)t)) = (k/m) * (A^2-(L2-L)^2) = ((F^2)/(km) - (k/m)(L2-L)^2). Finally we have
v = sqrt((F^2)/(km) - (k/m)(L2-L)^2).
Good stuff.
In post 2, lemiant used energy conservation to find the solution*.
In post 6, dan2915 used the equations for simple harmonic motion.
Let me add a third way: use Newton’s 2nd Law:
F = -kx =** Ma** = M*(dV/dt) = M*(dx/dt)(dV/dx) = MV(dV/dx)
=> MV(dV/dx) = -k*x
=> V*dV = -(k/M)xdx
Integrate the left side from V=0 to V=V[sub]2[/sub], and the right side from x=-F/k to x=(L-L[sub]2[/sub]), and solve for V[sub]2[/sub]
This Quiz was inspired by post 5 in this thread.
*neglecting the error in the final step
*
*
