# Pendulum Hitting Peg

by keasterling
Tags: hit, peg, pendulum, period
 P: 6 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.
 P: 6 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???
 P: 6 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?