Spring theory, when can this happen? If ever

  • Thread starter Thread starter flyingpig
  • Start date Start date
  • Tags Tags
    Spring Theory
AI Thread Summary
In a scenario involving an ideal massless spring on a frictionless surface, when the spring is compressed and released, it will oscillate between its maximum compression and maximum extension points. If a mass is attached to the spring, it will accelerate towards the equilibrium point, converting potential energy into kinetic energy, and will stop at the equilibrium position after passing through it. The spring itself, being massless, does not oscillate but remains at the equilibrium point once the mass reaches it. Without an attached mass, the spring would theoretically oscillate infinitely between its maximum compression and extension. Thus, the behavior of the spring depends significantly on whether a mass is attached.
flyingpig
Messages
2,574
Reaction score
1

Homework Statement



I have a spring on a horizontal flat frictionless surface. If I push it at its max at x = -x_max and releases it, will it stop at the equilbrium point and go back to x = -x_max or will it go beyond the x = 0 and go to x = x_max?

I know that if i have an object attched to the spring (but not becoming one with the spring) it will stop at x = 0 and releases it, but will it go back and forth between -x_max and x_max if there is no object?
 
Physics news on Phys.org
Is your spring a real spring or an ideal (massless) spring?
 
hmmm let's go with the ideal massless spring
 
I imagine that if you compress/stretch an ideal massless spring and then release it, it would snap back to its neutral position instantaneously, with no oscillation. EDIT: oops, I meant to say it will oscillate from x_min to x_max at infinite frequency.
 
Last edited:
What if there is an object? Will it just stop at x = 0?
 
Yes. Recapping the scenario: an ideal massless spring on a frictionless horizontal surface is fixed at one side, equilibrium length is x. The spring is compressed to a length of x-x_max, and a mass is placed next to the compressed spring. At this point in time, the potential energy in the spring is given by:

PE-x_max = (1/2)k(-x_max)2

As the mass-spring system is released the mass begins to accelerate towards the spring's equilibrium point, gaining kinetic energy. At the equilibrium point, there is no more PE in the spring. It has all been converted to the KE of the mass. The mass will continue moving, now at constant velocity. Since the spring is massless (no inertia) it will remain at the equilibrium point.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top