Physical Pendulum - II

A rod AB of length l = 12 2 l=12\sqrt 2 m whose linear density λ = λ 0 x l \lambda =\frac { { \lambda }_{ 0 }x }{ l } where λ 0 { \lambda }_{ 0 } is a positive constant and x x is distance from end A is pivoted at distance d d from the center of mass of the rod such that end A is above the the point where it is pivoted. Find the value of d d so that the time period of oscillation is minimum.

This problem is originally part of set Mechanics problems by Abhishek Sharma .
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1 solution

Jafar Badour
May 9, 2015

Nicely done. You could've saved time by using AM-GM inequality rather than differentiating.

Abhishek Sharma - 6 years, 1 month ago

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Wow.. i dont have so much of observational skills ... i didnt looked out for AM-GM. Its nothing but the side effect of our Sir.

Anything comes... He says.. "Take log and differentiate"

Md Zuhair - 3 years, 5 months ago

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That is very usual..Physics teachers usually prefer calculus. @Md Zuhair

Ankit Kumar Jain - 2 years, 10 months ago

I cannot understand the first point where you have written the expression for time period directly.

Atul Solanki - 5 years, 5 months ago

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