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A uniform square plate of mass m is supported in a horizontal plane by a vertical pin at B and is attached at A to a spring of constant K. If corner A is given a small displacement and released, determine the period of the resulting motion

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$2 π \sqrt{\frac{2 m}{3 k}}$

$\sqrt{\frac{m}{3 k}}$

$π \sqrt{\frac{m}{3 k}}$

$2 π \sqrt{\frac{m}{6 k}}$

answer is A.

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Detailed Solution

[Step 1]: $τ = I α$
[Step 2]: $F b = \frac{2}{3} m b^{2} α$
[Step 3]: $- k b x = \frac{2 m b^{2} α}{3}$
[Step 4]:

$\begin{array}{l} α = \frac{3 k x}{2 m b} \\ \Rightarrow \frac{a}{b} = \frac{3 k x}{2 m b} \\ \Rightarrow a = \frac{3 k x}{2 m} \\ c o m p a r i n g w i t h a = - ω^{2} x \\ w e g e t \\ ω^{2} = \frac{3 k}{2 m} \end{array}$

[Step 5]: $ω = \sqrt{\frac{3 k}{2 m}}$
[Step 6]: $T = 2 π \sqrt{\frac{2 m}{3 k}}$

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