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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 20:42:51 IST
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A particle moves in a circular path with uniform angular velocity. The displacement of the particle from a starting point can be shown on the graph as:
Options:
straight line triangle
semi circle
sinusoidal
parabolic
ALL WITH BASE AS +ve X-AXIS
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26 Mar 2008 20:43:45 IST
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c
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26 Mar 2008 20:54:56 IST
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(C)
Sinusoidal
Take in consideration projection of a particle in uniform circular motion on a screen on the ground.
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______________________________________________
Siddhant Shah
The trouble with doing something right the first time is that nobody appreciates how difficult it is.
You can't control the wind, but you can adjust your sails.
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26 Mar 2008 20:57:08 IST
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that's right (C).
But plz explain a little bit more.
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26 Mar 2008 21:01:44 IST
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draw the projection of the point at any instant on the x axis..
the distance of the projection from the point will be = R cos
but theta is nothing but  * t
so, position will be = R cos (omega*t)
which is sinusoidal in nature
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JEE and OLYMPIA INFINATUM
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26 Mar 2008 21:09:25 IST
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can you include an image to explain the answer plz.
showing where is R(cos)(theta).
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27 Mar 2008 01:22:11 IST
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sorry for the late reply mate.. had logged off
anyway, here's the image.. hope its useful :)
click on it to enlarge..
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JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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27 Mar 2008 05:07:14 IST
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consider a starting point as any point on the circle.
Now let the particle move around the circle... now at different positions see the length of the line joining the particle to the starting point... you will see that the length (displacement) is least , 0, when it reaches the starting point, and max when it is opposite diametrically across the starting point.... this process repeats continuously along its motion hence a sinusoidal curve
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* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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