Question:
does the length of the pendulum swing affect how long it takes for one swing?
Qwertyuiop
2011-09-01 04:24:10 UTC
i need help ASAP!
i am investigation the question above ^
i have already tested it, and found out the results, etc, and i have found out that yes, the length of the string of the pendulum effects the time it takes for one swing.
now i need to write my conclusion, and i have to state why this is... any ideas? i have already put because it has a longer distance to travel....
thanks in advance!
Four answers:
ZL777
2011-09-01 04:40:22 UTC
Yes the length determines the length for one swing. It doesent how much the object Weight is on the string. you could maybe incorperate this into your conclusion. For example ' no matter if you put a 5kg block to the string or a small bead, the swing time is mainly effected by the length.
justin
2011-09-01 04:29:05 UTC
It only effects the time of travel when the weighted end of the pendulum hasn't been increased in equal proportion to the rest of it. In other words, if one pendulum takes 2 secs to complete the swing, then if you make another one that exactly the same to the scale but 3 times bigger, it will still take 2 secs to complete the swing.



But one thing that might help some people better explain as to why the time is different when everything isn't increased the same amount- it would help to mark which kind of pendulum.
stanberry
2016-11-29 14:03:00 UTC
For the needs of this question, i'm going to examine "hight" and length of string because of the fact the comparable. i can't see what different sort of hight applies. in fact, this, length of string (or despite) is the only element that determines the frequency of the swings. a easy weight on the tip of the line will swing on the comparable value as a heavy one. The heavier weight would propose that much less will intervene with the passing of the pendulum, which includes wind resistance. comparable with mass. a huge, empty sphere will swing on the comparable value as a smaller, sturdy sphere of the comparable mass. back, this assumes no air resistance. None of this concerns to you, needless to say, except you comprehend what a pendulum is.
Kiran Iyer
2011-09-01 04:27:55 UTC
as the time period of the pendulum is given by:T = 2 pi square root{l/g}........from this relation it is proved.


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