Why is the Coefficient of cosθ Negative in the Exact Pendulum Solution?

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Homework Help Overview

The discussion revolves around the mathematical treatment of the coefficient of cosθ in the context of pendulum motion. Participants are examining the implications of the sign of this coefficient in relation to the behavior of the pendulum over time.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are exploring the necessity of a negative coefficient for cosθ and questioning the correctness of the book's expression compared to the original poster's derivation. There is also discussion about the behavior of θ over time and its implications for the choice of square root.

Discussion Status

The discussion is active, with participants providing insights and questioning assumptions. Some suggest that the original poster's derivation may be correct, while others are examining the implications of the sign of the square root in relation to the pendulum's motion.

Contextual Notes

There is mention of the condition that θ is usually less than or equal to θ0, which affects the realness of the square root. The discussion also touches on the timing of the pendulum's motion and how that relates to the signs used in the equations.

Saptarshi Sarkar
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Homework Statement
Show that the solution of the exact pendulum equation
can be written in the given form
Relevant Equations
##\ddot \theta = -\frac{g}{l}sin\theta##
##t = \int d\theta \frac {1}{\sqrt{A-\omega^2 cos\theta}}##
The question is

1583505635525.png


My attempt is given below

_.jpg


I am not sure what to do now. I don't understand how I can make the coefficient of ##cos\theta## negative.
 

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Hello Saptarshi,

I suspect you don't have to. Your derivation seems correct to me and agrees with what we see here

Moreover, usually ##\ \theta \le \theta_0\, , \ ## so ##\ \cos\theta \ge \cos\theta_0\ ## and the square root is real.

I think the book expression is wrong and yours is right.
 
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BvU said:
Hello Saptarshi,

I suspect you don't have to. Your derivation seems correct to me and agrees with what we see here

Moreover, usually ##\ \theta \le \theta_0\, , \ ## so ##\ \cos\theta \ge \cos\theta_0\ ## and the square root is real.

I think the book expression is wrong and yours is right.
In addition to this, I think the OP should have chosen the negative square root, rather than the positive. theta is decreasing with time.
 
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Chestermiller said:
theta is decreasing with time
Only half the time.
 
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Thanks. I guess I should have added "initially." However, the minus sign prevails at all the times.
 
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