# How to determine the range of time period for a simple pendulum

I have solved the first part of the problem. While solving the next part I have taken general oscillation(i.e. NOT small oscillations) I have obtained time period T as T=$$\ 2\pi(\sqrt(l/g))(1+(1/4)(sin^2(\alpha/2)))$$. However I am unable to proceed any further. How can I obtain the inequality?

• You may want to share this question on physics stack exchange as well. Here is the link: physics.stackexchange.com – Aniket Gupta Feb 21 at 17:34
• Since $\alpha = 30^\circ$, Your formula gives $$T = 2\pi\sqrt{\frac lg}\frac {17}{16}$$ but $17/16 = 1.0625$, which is not between $1$ and $\sqrt{\pi /3} \approx 1.0233$. So I suspect you made an error in figuring out $T$. But that isn't how you were supposed to solve this anyway. You just finished producing an inequality based on the acceleration in the variable $y$. The obvious intention here is that you should figure out $\ddot \theta$ for the pendulum and show that maximum value is $\frac gl$ and minimum value is $\frac {3g}{\pi l}$. – Paul Sinclair Feb 22 at 4:17
• I have corrected. It would be half the angle – Sharmi C Feb 22 at 5:27
• I have found out that $\ddot \theta = -\frac gl sin\theta$ and I guess that this is NOT case of small oscillations – Sharmi C Feb 22 at 5:39
• Now you need to find bounds for $\frac{\sin θ}{θ}$ on the interval $θ\in[-\frac\pi6,\frac\pi6]$. Note that the function is almost parabolic there. – Lutz Lehmann Feb 22 at 9:59