Question

If a uniform square plate of edge a is suspended
through a corner such that it is free to rotate in
vertical plane, then its time period of oscillation is
​

Answer 1

Explanation:

Given If a uniform square plate of edge a is suspended  through a corner such that it is free to rotate in  vertical plane, then its time period of oscillation is

• We need to find the time period of oscillation.
• So given a uniform square plate of edge a is suspended through a corner then t = a/√2.
• We know that time period T = 2π√ I / mgt
• Now I (centre of mass) = ma^2 / 6
• Now I (point of suspension) = ma^2 / 6 + ma^2 / 2
•                                            = 2ma^2 / 3
•          So T = 2π √2/3 ma^2 / mg a/√2

So T = 2π√2√2 a / 3g

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