Question

A simple pendulum of length L is constructed from a point object of mass m suspended by a massless string attached to a fixed point. A small peg is placed a distance 2L/3 directly below the fixed pivot point so that the pendulum would swing as shown in the figure below. The mass is displayed 5 degrees from the vertical and released. How long does it take to return to its starting position?

• A. $\pi \sqrt{\frac{L}{g}}(2+\frac{2}{\sqrt{3}})$
• B. $\pi \sqrt{\frac{L}{g}}(1+\frac{1}{3})$
• C. $\pi \sqrt{\frac{L}{g}}(1+\sqrt{3})$
• D. $\pi \sqrt{\frac{L}{g}}(1+\frac{1}{\sqrt{3}})$

Answer 1

The correct option is A $\pi \sqrt{\frac{L}{g}}(2+\frac{2}{\sqrt{3}})$

$\text{over the reaction}A-B\text{, the pendulim have length L}$

$\text{Time period}=2\pi \sqrt{\frac{L}{g}}$

$\text{Time from}A-B=\frac{1}{4}\times 2\pi \sqrt{\frac{L}{g}}$

$\text{over,}B-C-B\text{, time taken}=\frac{1}{2\pi }\times 2\pi \sqrt{\frac{L}{3g}}$

$\text{over,}B-A=\frac{1}{4}\times 2\pi \sqrt{\frac{L}{g}}$

$\text{total time}=\frac{1}{2}\times 2\pi \sqrt{\frac{L}{g}}+\frac{1}{2}\times 2\pi \sqrt{g^{L/3}}$

$=2\pi \sqrt{\frac{L}{g}}(1+\frac{1}{\sqrt{3}}]$

$=\frac{\sqrt{3}+1}{\sqrt{3}}\times 2\pi \sqrt{\frac{2}{8}}$

$=\pi \sqrt{\frac{L}{g}}[2+\frac{2}{\sqrt{3}}]$