Solution for the exact pendulum

[Saptarshi Sarkar]

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

My attempt is given below

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

 

[BvU]

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.

 

[Chestermiller]

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.

 

[haruspex]

theta is decreasing with time

Only half the time.

 

[Chestermiller]

Thanks. I guess I should have added "initially." However, the minus sign prevails at all the times.