Find the point of separation in SHM

[Vivek98phyboy]

In the given problem, i can understand that after placing the two blocks in equilibrium it oscillates with an amplitude of

(2/k)(m1+m2)gsinø

The answer for (b) is given as

(1/k)(m1+m2)gsinø

To my knowledge, m2 separate from m1 when the acceleration is greater than gsinø and so they should be separating only at max displacement on the right side as the m1 retraces its path after imparting the greatest acceleration to m2. So shouldn't the answer be

(2/k)(m1+m2)gsinø

on the right? Where am i wrong?

 

[jbriggs444]

The answer for (b) is given as To my knowledge, m2 separate from m1 when the acceleration is greater than gsinø and so they should be separating only at max displacement

What would be the downward acceleration at the position of max displacement? (where both the spring and gravity are pulling downward?

 

[DEvens]

Maybe if you think about acceleration as opposed to velocity, you will see the problem. The acceleration of the mass m1 is not always in the same direction as the velocity. Also, what is the condition that the two blocks separate?

 

[Vivek98phyboy]

Maybe if you think about acceleration as opposed to velocity, you will see the problem. The acceleration of the mass m1 is not always in the same direction as the velocity. Also, what is the condition that the two blocks separate?

They separate when acceleration of the spring is greater than gsinø

 

[DEvens]

They separate when acceleration of the spring is greater than gsinø

So, where does that happen? Then keep going.

 

[Vivek98phyboy]

That happens when it is at 1/3 of amplitude to the right. I was able to find that but my doubt is that the acceleration is max. when it is at the max. Displacement. So the acceleration at the most extreme compression point itself should be greater than the acceleration at 1/3 of amplitude. If it is so, shouldn't the m2 leave m1 at the extreme position itself as the acceleration is already greater than gsinø?

 

[Vivek98phyboy]

What would be the downward acceleration at the position of max displacement? (where both the spring and gravity are pulling downward?

I think m1 will experience

[(omega)²(2/k)(m1+m2)gsinø]+[gsinø]

whereas m2 will experience

gsinø

 

[jbriggs444]

I think m1 will experience whereas m2 will experience

That would suggest that their accelerations are significantly different. They've already separated.

When do they first begin to separate?

 

[Vivek98phyboy]

That would suggest that their accelerations are significantly different. They've already separated.

When do they first begin to separate?

But how do you tell that?

 

[jbriggs444]

But how do you tell that?

What happens if the spring is pushing on the lower block?
What happens if the spring is pulling on the lower block?
What happens if the spring is neither pushing nor pulling on the lower block?

 

[Vivek98phyboy]

What happens if the spring is pushing on the lower block?
What happens if the spring is pulling on the lower block?
What happens if the spring is neither pushing nor pulling on the lower block?

Got it. You say so because m2 is experiencing only gsinø, but it is only at that moment it separated as per the solutions. I tried checking it with various platforms and the solution was right. Problem is my intuition clashes with the displacement. They say it separates when it is as 1/k/k(m1+m2)gsinø but the acceleration at extreme position is bigger than the acceleration at any instant. Hence the should've separated at the extreme position itself. But I don't see so here

 

[jbriggs444]

Got it. You say so because m2 is experiencing only gsinø, but it is only at that moment it separated as per the solutions. I tried checking it with various platforms and the solution was right. Problem is my intuition clashes with the displacement. They say it separates when it is as 1/k/k(m1+m2)gsinø but the acceleration at extreme position is bigger than the acceleration at any instant. Hence the should've separated at the extreme position itself. But I don't see so here

They should separate the instant that the spring starts tugging on the lower mass instead of pushing. No need to wait for it to tug maximally hard.

 

[Vivek98phyboy]

They should separate the instant that the spring starts tugging on the lower mass instead of pushing. No need to wait for it to tug maximally hard.

Yeah. Now i am able to think of it properly. It makes sense when it separates after passing the equilibrium point. Another doubt, will you use the term amplitude on any of these distances ? As they are separated, it is no more a SHM. The block just jumps up and down insanely after this, so i think the usage of term amplitude in this problem isn't logical. Am i right?

 

[jbriggs444]

Yeah. Now i am able to think of it properly. It makes sense when it separates after passing the equilibrium point.

It is not the equilibrium point. At the equilibrium point, the spring is pushing hard enough to hold both blocks in place. Or to give them zero net acceleration. The point of separation is the point where the spring is exerting zero force -- the point where the spring is at its relaxed length.

Another doubt, will you use the term amplitude on any of these distances ? As they are separated, it is no more a SHM. The block just jumps up and down insanely after this, so i think the usage of term amplitude in this problem isn't logical. Am i right?

You could use the term "amplitude" even though the behavior only goes through a portion of a cycle. Though I would use some extra words to refer to "the point where peak upward displacement would be if the blocks were glued together".

 

[Vivek98phyboy]

Okay. So that's why we got

It is not the equilibrium point. At the equilibrium point, the spring is pushing hard enough to hold both blocks in place. Or to give them zero net acceleration. The point of separation is the point where the spring is exerting zero force -- the point where the spring is at its relaxed length.

You could use the term "amplitude" even though the behavior only goes through a portion of a cycle. Though I would use some extra words to refer to "the point where peak upward displacement would be if the blocks were glued together".

Okay. So that's why we got the answer as 1/k(m1+m2)gsinø away from the equilibrium position ( 3/k(m1+m2)gsinø was the total length comprrssed from unloaded state).
Here we have an inclined scenario. What if i keep the spring vertical and repeat the same thing with some arbitrary measurements, i hope it will get released at natural length of the spring? Is it right?

 

[jbriggs444]

Here we have an inclined scenario. What if i keep the spring vertical and repeat the same thing with some arbitrary measurements, i hope it will get released at natural length of the spring? Is it right?

We have solved the problem for an arbitrary angle $\theta$. So plug in $\theta = 90$ degrees.

$$\frac{m_1+m_2}{k}g sin \theta = \frac{g(m_1+m_2)}{k}$$

Which, by no coincidence, is the relaxed position. So yes, you are right.