1
$\begingroup$

enter image description here

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?

$\endgroup$
  • $\begingroup$ You may want to share this question on physics stack exchange as well. Here is the link: physics.stackexchange.com $\endgroup$ – Aniket Gupta Feb 21 at 17:34
  • $\begingroup$ 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 $\frac {3g}{\pi l}$. $\endgroup$ – Paul Sinclair Feb 22 at 4:17
  • $\begingroup$ I have corrected. It would be half the angle $\endgroup$ – Sharmi C Feb 22 at 5:27
  • $\begingroup$ I have found out that $\ddot \theta = -\frac gl sin\theta$ and I guess that this is NOT case of small oscillations $\endgroup$ – Sharmi C Feb 22 at 5:39
  • $\begingroup$ 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. $\endgroup$ – Lutz Lehmann Feb 22 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.