Solving Spring Confusion: Ee2 Canceled Out or Not?

[FarisL]

So I got this question on a physics quiz:

"You have have two masses, M₁ and M₂, weighing 1.00 kg and 2.00 kg respectively. You have an ideal, massless spring (spring constant = 40.0 N/m) set up vertically from the ground with a massless platform attached to its free end on top. The two masses are gently placed on top of each other on the platform, so that the mass-spring system is at rest, with the platform's height set as zero relative to the system (for simpler calculations). Suddenly, the 2.00 kg mass is removed, causing the system to rebound vertically upwards. Find the maximum height the platform reaches."

You are first required to find the distance, x, the spring contracts when the two masses are placed on it, using kx = mg. Once you know that, you are to find the maximum height using the equation for the conservation of energy, E_k1+E_g1+E_e1 = E_k2+E_g2+E_e2, where E_k is the kinetic energy, E_g is the gravitational energy, and E_e is the elastic energy.

Now, both E_k1 and E_k2 are canceled out, as they equal zero. E_g1 is canceled out as well, as the height at that moment is zero as well. We are left with E_e1 = E_g2+E_e2.

My question: Is E_e2 canceled out or not?
I solved it using the two scenarios; if it is, you should get an answer of around 1.1 m. If not, you should get something like 0.981 m.

 

[AlephZero]

My question: Is E_e2 canceled out or not?

No. Think what would happen if M2 was very small (say 1 gram). The spring would rebound a small distance, but not far enough to release all its strain energy.

 

[FarisL]

But then there's the argument that, for a system to reach its maximum height, all the energy must be concentrated as gravitational potential energy. In order to do that, both kinetic and elastic energy must equal zero, according to the law of conservation of energy; hence E_e2 is canceled out.

That's why I'm confused. To me, the concepts seem to contradict each other.

 

[AlephZero]

But then there's the argument that, for a system to reach its maximum height, all the energy must be concentrated as gravitational potential energy.

That is not correct. When the velocity is zero, all the energy must be in some form of potential energy, but there is no reason why it must all be gravitational potential energy.

Think about a mass on a spring vibrating horizontally. There is no change in gravitational PE. The energy is transferred from all kinetic energy when the spring is at its unstretched length, to all elastic potential energy when spring is at its maximum length and the velocity is zero.

 

[FarisL]

Thing is, like I said before, this came up on my quiz. My teacher firmly believes that everything but the gravitational energy is canceled out, since if you were to look at it from a mathematical view of point, the height would end up maximum. Though your reasoning makes perfect sense. I'll have to ask him once more.

Thank you for your replies

 

[vanhees71]

I'd rather analyze the forces and solve the equations of motion.

First, when both masses are sitting on the spring, we have given, that the whole thing is at rest, i.e., the forces are $$0$$. Let $$x=0$$ be the end of the spring, if there's no mass on it. Now, with both masses sitting on the spring, this end is at $$x=x_0<0$$. Vanishing of the total force on these masses says:

$$-D x_0-(M_1+M_2)g=0 \; \Rightarrow \; x_0=-\frac{(M_1+M_2)g}{D}$$

When now the mass, $$M_2$$, is taken suddenly away, we have the initial condition for the remaining mass $$M_1$$:

$$x(0)=x_0, \quad \dot{x}(0)=0$$.

Now the equation of motion reads

$$M_1 \ddot{x}=-D x-M_1 g$$

The general solution of this equation is

$$x(t)=A \cos(\omega t)+B \sin(\omega t)-\frac{M_1 g}{D}$$

with $$\omega=\sqrt{D/M_1}$$.

With the initial conditions, we get

$$x(t)=\left (x_0+\frac{M_1 g}{D} \right ) \cos(\omega t)-\frac{M_1 g}{D}$$

or

$$x(t)=-\frac{M_2 g}{D} \cos(\omega t)-\frac{M_1 g}{D}$$

The maximum height is reached for $$\cos(\omega t)=-1$$, i.e., at $$t=T/2=\frac{\pi}{\omega}$$. The height obviously is

$$x_{\text{max}}=\frac{(M_2-M_1) g}{D}$$.

 

[FarisL]

Wait, what does D represent?

 

[AlephZero]

In the other post D is the spring stiffness. English math books usually use K.

You can see the previous answer is right by a simpler argument.

With both masses on the spring, the equilibrium position is

$$x_0 = -(M_1 + M_2)g/K$$

With one mass on the spring, the equilibrium position is

$$x_1 = -M_1g/K$$

If you release the spring at $x_0$ and its equilibrium position is $x_1$, it will do simple harmonic motion about $x_1$ with amplitude $x_0-x_1$.

So the highest and lowest points of the motion must be $x_1 - (x_0-x_1)$ and $x_1 + (x_0-x_1)$, or $x_0$ and $2x_1-x_0$.

$$2x_1 - x_0 = -(M_1 - M_2)g/K$$

 

[AlephZero]

Ahhhh...

I think I see what your teacher is saying now, but I don't like his explanation for it.

The math in posts #6 and #8 assumes that the masses are fastened to the platform, not just resting on it. With that assumption it is possible for the maximum height to be above the unstretched length of the spriing, depending on the values of M1 and M2. This would happen for the values of M1 and M2 in your OP.

When the platform is above the unstretched length of the spring, the spring is pulling down on the mass, but the inertia of the mass and its upwards velocity mean it will rise to a higher position before it stops.

But in the question the mass is not fastened to the platform, it is resting on it. So when spring reaches its unstretched length, the platform will stop moving upwards and the mass will leave the platform and move higher.

This is not about energy, it is about the fact that the force between the platform and the mass can only act one way.

As I said before, if M2 is small compared with M1, the platform will not return to its original position. The math agrees with that statement.