Solving for Period of SHM with Pendulum Hitting Peg

[keasterling]

1. A simple (massless string, point mass pendulum) pendulum swings down and hits a peg located at the equilibrium line halfway through the pendulum's swing. The pendulum's string has a length L when it's to the left of the peg, and length \frac{L}{2} after it hits the peg.

(a)Is this simple harmonic motion? Even approximately simple harmonic?
(b) What is the period of the motion?

2. I say yes, when the pendulum is released, it will hit the peg, swing up to the right to the same height it was released from on the left side, swing back down to the the left, and repeat the cycle all over again with a definite period. Thus, SHM... I guess.

As for (b), I know you have to use conservation of energy, but I'm not quite sure how to translate that type of work into the formulas for period. Any help/hints would be appreciated. Not looking for a direct answer, just a nudge in the right direction.

 

[turin]

(a) is testing you on two points: (i) the definition of SHM, and (ii) recognizing the harmonic structure of the pendulum. You have failed (i).

For (b), I can only assume that you are expected to assume small angular variations, so that the system is approximately linearized. (in general, a pendulum that swings at large angles will not obey the formula for frequency that you have probably been taught). Anyway, you must realize that the pendulum has, effectively, two moment arms, and then you must reason on the basis of symmetry how to modify the formula for frequency.

 

[keasterling]

Yeah, I looked into part (a) a bit more, and I guess that it really isn't SHM mainly because the graph of position would not be sinusoidal. In other words, the amplitude of the graph would not be equal on either side because the pendulum wouldn't be moving equal distances from equilibrium on each side. Simple pendulum motion on it's own wouldn't be SHM anyway, unless you consider small angles of \Theta where sin(\Theta)=\Theta. So you're totally right on. Anybody see a problem with that one?

For a simple pendulum application (massless string, point mass pendulum), the period is defined as T=2\pi\sqrt{\frac{L}{g}}where L is... L (length of the pendulum), and g is the good old 9.8.

So... I'm not quite sure how to go onwards with (b) from here. Even though the movement isn't SHM, it still has a definite period. The answer will be entirely symbolic as well, I just don't know how to get there. Once again, I'm feelin' a bit of conservation of energy coming on. Anybody?

 

[keasterling]

Maybe if I post again, people will come.

?

 

[turin]

I'm sorry. For some reason I didn't see this subscription update.

You are spot on with (a).

For (b), is L a constant throughout the period?

 

[keasterling]

Kind of. L is the overall length of the string (pendulum). The pendulum is swinging from left to right, and it hits a peg halfway through it's arc (positioned at the equilibrium). For the rest of the swing, the line has been shortened to L/2.

 

[keasterling]

Try this one on for size:

T=\pi\sqrt{\frac{L}{G}}+\pi\sqrt{\frac{L}{G}}

T=\pi\sqrt{\frac{L}{G}}(1+\sqrt{}\frac{1}{2}

How about that?
Yay? Nay?

 

[turin]

Why did you change to capital G? Be careful with that; it means something else, even in gravitation. Are you saying that you have two different periods?

 

[keasterling]

Yeah, I meant little "g," not the "6.67 x10 to the blah blah blah." My bad. There's not really two periods, just that the one period was divided up into two sections. One half of the period was when the string length was just L. The next half of the period is for when the length is (L/2). And... you divide the two portions by two, so it's just "pi*sqrt(L/g) + pi*sqrt(L/2g)" I guess I typed it up wrong in my previous post. How does that sound?

 

[turin]

OK. So how do you think you should combine these two results to get the actual total period? Hint: compared to the corresponding simple pendulum, for what fraction of the period does the pendulum actually behave that way?