Physics Help Forum Pendulum

 Oct 19th 2008, 09:14 AM #1 Member   Join Date: Jul 2008 Posts: 60 Pendulum Hello. I have this Pendulum, I need to find the time that it takes it to go from A to B. I though about using the good old Kinematics formula v = v0 + at, finding v through energy equation, but then I realized that a constantly changing, so, can I use it? If not, is there another way? 10x. Attached Thumbnails
 Oct 19th 2008, 11:13 AM #2 Senior Member     Join Date: Apr 2008 Posts: 815 There are several formulas about the case of a pendulum. One of them is that $\displaystyle T=2\pi \sqrt{\frac{l}{g}}$, where $\displaystyle T$ is the period of the pendulum, $\displaystyle l$ is the longitude of the pendulum and $\displaystyle g$ is the gravitational acceleration. In your sketch, we see that point A from point B is a quarter of a period. That is, the time taken for the pendulum to go from point A to point B is $\displaystyle \frac{T}{4}s$. Or, equivalently $\displaystyle \frac{\pi}{2}\cdot \sqrt{\frac{l}{g}}s$. __________________ Isaac If the problem is too hard just let the Universe solve it. Last edited by arbolis; Oct 19th 2008 at 11:16 AM.
Oct 19th 2008, 12:52 PM   #3
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 Originally Posted by arbolis There are several formulas about the case of a pendulum. One of them is that $\displaystyle T=2\pi \sqrt{\frac{l}{g}}$, where $\displaystyle T$ is the period of the pendulum, $\displaystyle l$ is the longitude of the pendulum and $\displaystyle g$ is the gravitational acceleration. In your sketch, we see that point A from point B is a quarter of a period. That is, the time taken for the pendulum to go from point A to point B is $\displaystyle \frac{T}{4}s$. Or, equivalently $\displaystyle \frac{\pi}{2}\cdot \sqrt{\frac{l}{g}}s$.
How do you know it's a quarter of a period?

10x.

 Oct 19th 2008, 01:02 PM #4 Senior Member     Join Date: Apr 2008 Posts: 815 Because a period is when the little ball of the pendulum reaches the same place with the same vector velocity and same vector acceleration. If you are at point A and let the ball "fall off", it will pass by point B, reaches a maximum height and then fall off again but in the opposite direction to reach another time point B and finally point A once again. The time required for that is precisely a period. But you could ask "but it pass twice by point B, so why it hasn't made a period?", well it's because the first time it passes by point B it has an opposite vector acceleration and velocity as when it passes the second time. So half a period is when the ball reaches a maximum in height (same height as point A, but on the left on your sketch. Divide it by 2 again and you see that it is where B is placed. So a quarter of a period is at point B, if you start from point A in your sketch. __________________ Isaac If the problem is too hard just let the Universe solve it.
Oct 19th 2008, 01:04 PM   #5
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 Originally Posted by arbolis Because a period is when the little ball of the pendulum reaches the same place with the same vector velocity and same vector acceleration. If you are at point A and let the ball "fall off", it will pass by point B, reaches a maximum height and then fall off again but in the opposite direction to reach another time point B and finally point A once again. The time required for that is precisely a period. But you could ask "but it pass twice by point B, so why it hasn't made a period?", well it's because the first time it passes by point B it has an opposite vector acceleration and velocity as when it passes the second time. So half a period is when the ball reaches a maximum in height (same height as point A, but on the left on your sketch. Divide it by 2 again and you see that it is where B is placed. So a quarter of a period is at point B, if you start from point A in your sketch.

Got you, 10x.

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