How does trigonometry help determine the gravity component on an inclined plane?

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jaewonjung
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Homework Statement
Question is in the file.
Relevant Equations
T=2pi root(l/g)
Since gravity is acting downward, I found the gravity component parallel to the plane, which was g/sin60.
I substituted g/sin60 for g into that equation and got D, but the answer should be C.
 

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kuruman said:
Look at it this way. Which of the choices (1) depends on the angle and gives the expected answer when the angle of the incline is, instead of 60o, (2) 90o and (3) 0o ?
Well, I know that A is wrong since it is independent of the angle. The other answer choices have an angle component.
 
In these sorts of MCQs, it saves time to think of the extremes - what happens if the plane were absolutely horizontal or completely vertical? What would the period be in those cases? Which of of the choices make sense in that light? You'll have your answer.
 
hurreechunder said:
In these sorts of MCQs, it saves time to think of the extremes - what happens if the plane were absolutely horizontal or completely vertical? What would the period be in those cases? Which of of the choices make sense in that light? You'll have your answer.
Thanks for the reply!

I see. If the angle was approaching 0, the period would approach infinity, since the pendulum would almost be completely horizontal and gravity would have a negligible effect. I understand why C is the answer and not D, but I still don't understand how to algebraically find the equation for the period.
 
Since the plane is frictionless, only the component of gravity along the plane matters, which is gsin60. It is then identical to the case of a simple pendulum where g is replaced by g sin60
 
I found the gravity component parallel to the plane, which was g/sin60.
Review your trigonometry that lead you to that conclusion.