Simple Harmonic Motion of Meterstick

[ln(]

Homework Statement

Homework Equations

$\tau = rFsin(\theta)$
$\tau_{net} = I\alpha$
$F = -kx$
$kx = mg$

The Attempt at a Solution

I don't understand how the restoring force from the bending of the ruler behaves (so I have no idea how to apply torque here). I also don't understand how to quantify the "barely loses contact" condition. However, if I just ignore all that and pretend it behaves exactly like a spring and solve for the equilibrium position, I somehow arrive at the correct answer (A) using the last two relevant equations.

I assumed that $T = 2\pi \sqrt{m/k}$, where $T = 0.5 ~s$ as given. I get a relationship between $m$ and $k$ and then plugged this into $kx = mg$ to find that $x = x_{equilibrium} \approx 6.2 ~cm$.

Can someone explain how to actually do this problem and why what I did even works at all?

 

[haruspex]

why what I did even works

In effect, you have treated the ruler as though it is a spring with a spring constant of the right value to produce the known frequency when loaded with the small mass. In reality, all the mass of relevance to the oscillation is in the ruler, but your model is equivalent and simpler.

Now, suppose you were to place the mass carefully on top of the relaxed spring and let go. If it descends x to the equilibrium position, how much further will it descend before bouncing back up?
Where will be the top of the oscillation? What will be the acceleration of the mass at the top of the oscillation?

 

[ln(]

In effect, you have treated the ruler as though it is a spring with a spring constant of the right value to produce the known frequency when loaded with the small mass. In reality, all the mass of relevance to the oscillation is in the ruler, but your model is equivalent and simpler.

Now, suppose you were to place the mass carefully on top of the relaxed spring and let go. If it descends x to the equilibrium position, how much further will it descend before bouncing back up?
Where will be the top of the oscillation? What will be the acceleration of the mass at the top of the oscillation?

The equilibrium position here isn't parallel to the table, correct? Because the ruler has mass itself and the equilibrium position should be be at some angle below the horizontal. Just checking my understanding.
So then, wouldn't it descend $x$ below the equilibrium before bobbing back up?
The top of the oscillation should be $x$ above the equilibrium.
The acceleration at the top is just $kx/m$.

 

[haruspex]

The equilibrium position here isn't parallel to the table, correct?

Right.

wouldn't it descend x below the equilibrium before bobbing back up?

Yes.

The top of the oscillation should be x above the equilibrium.

Yes, back at the initial position.

The acceleration at the top is just kx/m

Well, yes, but that's not the simplest or most obvious form, which makes me wonder how you arrived at that.
Remember, it is back in its initial position and the spring is relaxed now.

 

[ln(]

Right.

Yes.

Yes, back at the initial position.

Well, yes, but that's not the simplest or most obvious form, which makes me wonder how you arrived at that.
Remember, it is back in its initial position and the spring is relaxed now.

Well then it would also be $g$ wouldn't it?

 

[PeroK]

Well then it would also be $g$ wouldn't it?

I would focus on the SHM. Is the "spring constant" important? What key quantity are you given in the problem statement?

 

[ln(]

I would focus on the SHM. Is the "spring constant" important? What key quantity are you given in the problem statement?

The period, of course.
So now I think we've established that a simple spring system can be used here as a model, since it says SHM. But I still have no quantitative understanding of the "barely loses contact" condition.

 

[haruspex]

Well then it would also be $g$ wouldn't it?

Yes.
What does that tell you about the normal force between the mass and the ruler at that point?

 

[ln(]

Yes.
What does that tell you about the normal force between the mass and the ruler at that point?

$F_N = 0$?

 

[haruspex]

$F_N = 0$?

Right. So is it on the point of losing contact?

 

[ln(]

Right. So is it on the point of losing contact?

Yes. But how is there a "minimum initial displacement" for that?
Wouldn't it lose contact if the mass has non-zero velocity at the top of the oscillation?

 

[PeroK]

Yes. But how is there a "minimum initial displacement" for that?
Wouldn't it lose contact if the mass has non-zero velocity at the top of the oscillation?

If you are in an elevator, going down say, what condition would cause your feet to come off the floor?

 

[ln(]

If you are in an elevator, going down say, what condition would cause your feet to come off the floor?

The elevator accelerating downwards with $a > g$.

 

[PeroK]

The elevator accelerating downwards with $a > g$.

Exactly.

 

[haruspex]

Yes. But how is there a "minimum initial displacement" for that?
Wouldn't it lose contact if the mass has non-zero velocity at the top of the oscillation?

In the model I described in post #2, placing the mass on top of a relaxed spring and letting go, yes it will always be true that at the top of the subsequent oscillations it will be on the point of losing contact.
But remember that this model only fits the question because of the way you calculated x. With a smaller displacement of the stick, the top of the oscillation would be (in the model) with the spring still partly depressed.

 

[ln(]

In the model I described in post #2, placing the mass on top of a relaxed spring and letting go, yes it will always be true that at the top of the subsequent oscillations it will be on the point of losing contact.
But remember that this model only fits the question because of the way you calculated x. With a smaller displacement of the stick, the top of the oscillation would be (in the model) with the spring still partly depressed.

The question says that the mass-ruler system is released after being pulled down by a distance $A$. I understand how the model in post #2 works, but the mass isn't simply being released here.
I also don't understand what you mean with your second point. My understanding is that if you don't simply release the mass onto the relaxed spring, but also pull it down initially, the top of the oscillation would certainly not be at the top (as in when the spring is relaxed).
Finally, I get that barely losing contact means $F_N = 0$ but I don't see how there is a minimum initial displacement associated with this. If you drag the mass-spring system down to its equilibrium where $kx = mg$, doesn't the spring not even oscillate?

 

[PeroK]

As I read the problem, the mass is very small, hence negligible. The ruler's oscillation is assumed to be unaffected by the mass. With this assumption I get one of the multiple choice answers.

 

[haruspex]

The question says that the mass-ruler system is released after being pulled down by a distance $A$. I understand how the model in post #2 works, but the mass isn't simply being released here.
I also don't understand what you mean with your second point. My understanding is that if you don't simply release the mass onto the relaxed spring, but also pull it down initially, the top of the oscillation would certainly not be at the top (as in when the spring is relaxed).
Finally, I get that barely losing contact means $F_N = 0$ but I don't see how there is a minimum initial displacement associated with this. If you drag the mass-spring system down to its equilibrium where $kx = mg$, doesn't the spring not even oscillate?

It is easy to get confused mapping between the actual question and the model.
As you answered in post #3, in the model, the mass would descend x below the equilbrium before bouncing back up. That corresponds to the initial displacement x, below equilibrium, of the meter stick in the question.
Equilibrium in the model is the spring depressed by x. Equlibrium for the meter stick is the stick sagging, stationary, under its own weight - the weight of the mass being negligible.
In the model, the mass has, well, mass, but the spring does not. For the stick, the stick has mass but the mass is, um, massless.

 

[haruspex]

I get one of the multiple choice answers.

As I read post #1, ln( got the right answer but does not understand why the method worked. It turns out to be a very neat method, but I agree it is not clear why it works, so I have been concentrating on explaining that.
ln( may also be interested in how to solve it with the more direct approach, but I haven't looked at that.

 

[ln(]

It is easy to get confused mapping between the actual question and the model.
As you answered in post #3, in the model, the mass would descend x below the equilbrium before bouncing back up. That corresponds to the initial displacement x, below equilibrium, of the meter stick in the question.
Equilibrium in the model is the spring depressed by x. Equlibrium for the meter stick is the stick sagging, stationary, under its own weight - the weight of the mass being negligible.
In the model, the mass has, well, mass, but the spring does not. For the stick, the stick has mass but the mass is, um, massless.

I see. So then we are looking for the top of the oscillation to see where the "barely loses contact" condition is relevant. In the model, this is $x$ above the equilibrium. For the ruler, this corresponds to $A$ above the equilibrium. And I know how to solve for $x = mg/k$, and this happens to be $A$. I do not see why this is the case, though. Why is there ONE particular (minimum) $A$ wherein the top of the oscillation is when contact is barely lost? I don't see how any smaller $A$ than 6.2 cm wouldn't allow this to be the case.
As a point of clarification: in the model, is the mass actually affixed to the spring? Because then I can see a relation between "barely losing contact" and the model. If you displaced the spring at all, instead of just plopping the mass, affixed, on top, then it wouldn't come up to the same unstretched position at the top of the oscillation. So then the model wouldn't really correspond to the problem, as you were saying. This gives me justification for the model used but I still don't see how the ruler itself needs to have a minimum initial displacement for the condition to apply.
Also, we must assume that the small mass doesn't alter the oscillations but still causes oscillations when plopped onto the ruler at rest/equilibrium?

 

[PeroK]

When I was at school we would fire various things about the classroom using our rulers as a slingshot. That's the idea here.

If you force the ruler far enough down it will project the mass up into the air.

If you can't see that you must have been too well behaved as a child!

 

[ln(]

When I was at school we would fire various things about the classroom using our rulers as a slingshot. That's the idea here.

If you force the ruler far enough down it will project the mass up into the air.

If you can't see that you must have been too well behaved as a child!

So then shouldn't the question ask for a maximum displacement rather than a minimum? If pushing the ruler down too much projects the mass up into the air.

 

[PeroK]

So then shouldn't the question ask for a maximum displacement rather than a minimum? If pushing the ruler down too much projects the mass up into the air.

No. There's a minimum displacement that results in projectile motion. The greater the displacement the further the object is projected. The max displacement is when the ruler breaks.

That said, it's not really a max or a min, it's the point at which the mass just leaves the stick.