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

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SUMMARY

The discussion centers on the negative coefficient of cosθ in the exact solution of a pendulum's motion. Participants agree that the original derivation by the user, Saptarshi, is correct and aligns with the expected behavior of the pendulum, where θ typically remains less than or equal to θ0, ensuring that cosθ remains non-negative. The consensus suggests that the textbook expression may be incorrect, and the negative square root should be utilized since θ decreases over time, reinforcing the need for clarity in the derivation process.

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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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