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

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Solution

R.E.F image
over the section
$A - B,$ the pendulum
have length L
Time period $= 2 π \sqrt{\frac{L}{g}}$
Time from $A - B = \frac{1}{4} \times 2 π \sqrt{\frac{L}{g}}$
over $B - C - B,$ time taken $= \frac{1}{2} \times 2 π \sqrt{\frac{L}{3 g}}$
Over $B - A = \frac{1}{4} \times 2 π \sqrt{\frac{L}{g}}$
total time $= \frac{1}{2} \times 2 π \sqrt{\frac{L}{g}} + \frac{1}{2} \times 2 π \sqrt{g^{L / 3}} = 2 π \sqrt{\frac{L}{g}} (1 + \frac{1}{\sqrt{3}})$
$= \frac{\sqrt{3} + 1}{\sqrt{3}} \times 2 π \sqrt{\frac{2}{8}}$

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R.E.F imageover the sectionA−B, the pendulumhave length L Time period =2π√LgTime from A−B=14×2π√Lgover B−C−B, time taken =12×2π√L3gOver B−A=14×2π√Lgtotal time =12×2π√Lg+12×2π√gL/3=2π√Lg(1+1√3)=√3+1√3×2π√28