Having difficulty with this Oscillation problem

[Theelectricchild]

This question is quite interesting:

A uniform meter stick of mass M is pivoted on a hinge at one end and held horizontal by a spring with spring constant k attached at the other end. If the stick oscillates up and down slightly, what is it's frequency? The length of the stick is one meter.

My thoughts: I believe I should write a torque equation about the hinge, but I am having difficulty doing that--- any suggestions?

Also and I f=[(k/M)^.5]/2pi but the back of the book has 3k where I have k. I don't understand how they get this. Thanks a lot for your help!

Also if you need a picture ill be happy to draw one out --- thanks again.

 

[arildno]

Clearly, as the rod rotates an angle [tex]\phi[/tex] about the hinge, there is a slight horizontal displacement of the rod (and spring) as well.
In the following, I neglect this displacement component, and model the problem with a pure vertical displacement.
If the displacement angle is [tex]\phi[/tex] then the associated vertical displacement of the tip of the rod (fixed to the spring) is [tex]L\phi[/tex] where L is the length of the rod (to the order of accuracy indicated).

The moment of inertia about the hinge must be: [tex]I=\frac{ML^{2}}{12}[/tex]
We gain thus:
[tex]kL^{2}\phi+\frac{ML^{2}}{12}\frac{d^{2}\phi}{dt^{2}}=0\to\omega^{2}=12\frac{k}{M}[/tex]

Since [tex]f=\frac{\omega}{2\pi}[/tex] we have:
[tex]f=\frac{\sqrt{\frac{3k}{M}}}{\pi}[/tex]

The factor 2 in the denominator in the book is wrong.

 

[Theelectricchild]

Ahh thank you much I understand all--- I know it's essential to solve the problem but WHY exactly do we use the [tex]kL^{2}\phi[/tex] ?--- or I should ask--- why is it "squared"? I am always quite good at solving and getting the correct answer, but not understanding why equations work can be quite annoying.

Thank you.

 

[Theelectricchild]

Uh oh i may have found a problem:

if you're just using (1/12)M*L^2, wouldn't you be saying that the axis of rotation WAS at the hinge which is at the END of the rod because you're using L which is the length of the rod? so wouldn't you use the (1/3)M*L^2? can you explain this one to me, ald.?

Interesting because if we use 1/3 then we end up getting the answer in the back which COULD be right?

Thanks!

 

[arildno]

In vector form, the spring force is [tex]\vec{F}=-kL\phi\vec{j}[/tex]
The arm from the hinge to the tip is, to same order of accuracy: [tex]\vec{r}=L\vec{i}[/tex]

Hence, we get the torque:[tex]\vec{r}\times\vec{F}=-kL^{2}\phi\vec{k}[/tex]

 

[arildno]

Uh oh i may have found a problem:

if you're just using (1/12)M*L^2, wouldn't you be saying that the axis of rotation WAS at the hinge which is at the END of the rod because you're using L which is the length of the rod? so wouldn't you use the (1/3)M*L^2? can you explain this one to me, ald.?

Interesting because if we use 1/3 then we end up getting the answer in the back which COULD be right?

Thanks!

You're right! Of course it should be 1/3 rather than 1/12. Sorry for that

 

[Theelectricchild]

Ahh thanks again! You help people out so much on this site and I do appreciate that.

 

[e(ho0n3]

I'm solving this exact same problem at the moment. Question: Why is the weight of the meter stick being ignored in the torque equation. Shouldn't the torque equation be:

[tex]\tau = -\frac{1}{2}LMg - kL^2\phi[/tex]

 

[arildno]

At a glance, I believe you're right.
Anyways (as I'm sure you're aware of), this wouldn't affect the frequency (which is what they want)