Solve Pendulum Problem: Find Time Period

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To solve the pendulum problem, the time period can be calculated using the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. The length provided is 40 cm, which converts to 0.4 m, and g is approximately 9.8 m/s². The angular amplitude of 0.04 radians is acceptable since it translates to about 2.29 degrees, well within the small angle approximation for simple harmonic motion. The calculated time period is approximately 1.26 seconds, confirming the solution is correct. The discussion emphasizes the importance of understanding the relationship between amplitude and the time period in pendulum motion.
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Homework Statement


This is not any school or any homework problem, It's just a problem I read in a book. And I don't know anything about it. So I would just like anyone to give me a brief guideline to solve this sum and please help me. (PS: don't tell me the answer, I want to try it on my own. Though I do need a lot of guidelines...)

I'm a real amateur, so you guys might find this sum easy, but i don't get it.
here it is:
A simple pendulum of length 40cm oscillates with an angular amplitude of 0.04 rad. Find the time period.


Homework Equations


T = 2 \pi \sqrt {L/g}
(that's all I know... I don't know about rad)


The Attempt at a Solution


Well I did think about it and came to a conclusion that since the time period doesn't depend on the amplitude of the pendulum well the time period will be:
T = 2 \pi \sqrt {L/g}
T = 2 * 3.14 \sqrt {0.4m/9.8}
therefore T = approx. 1.26s

Though I don't know what the answer is...
 
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The equation for the period of a simple pendulum is correct. However, it assumes that the amplitude of the pendulum's swing is not too great. It it gets too large, the motion is no longer simple harmonic. Most school books will allow an angle up to about 10 degrees as "small". The conversion of an angle in radians to degrees is given by 360 degrees is equivalent to 2 pi radians.
 
Umm...
Got it
when i converted the 0.04 rad into degree then I got it to be approx 2.29 degrees and since it is under 10 degrees as suggested by you my answer most probably is right and the motion is SHM.
Is it right?(I expect a yes though I will even accept a no too)
 
The answer is correct. :)
 
Yuhoooooooo!
Thanks for the help Stonebridge!
 
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