A bar magnet oscillates in a region of uniform magnetic field with time period T.  Now a piece of wood identical in all respect in terms of mass and length is joined to bar magnet as shown.  The now time period of oscillation is   

$\frac{I_1}{I_2}=\frac{1}{2}\Rightarrow T\alpha \sqrt{I}$

$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\left(\mathrm{\mu } \mathrm{B}\right)}}$

$T=2\mathrm{\pi }\sqrt{\frac{{\mathrm{I}}_1}{\left(\mathrm{\mu } \mathrm{B}\right)}} .........\left(1\right)$

$I_2=I_1+I_{wood}$

$$\begin{gathered} \sin ce the length and mass of the wood are identical to the bar magnet, \\ the moment of inertia of the wood\left(I_{wood}\right) will also be I_1. so \end{gathered}$$$I_2=I_1+I_1$

$Now time period with the wooden piece attached\left(T_2\right) will be$$T_2=2\mathrm{\pi }\sqrt{\frac{{\mathrm{I}}_2}{\left(\mathrm{\mu } \mathrm{B}\right)}}$

$T_2=2\mathrm{\pi }\sqrt{\frac{2I_1}{\left(\mathrm{\mu } \mathrm{B}\right)}}........\left(2\right)$

$from \left(1\right) and \left(2\right)$

$\frac{T_2}{T}=\frac{2\mathrm{\pi }\sqrt{{\displaystyle \frac{2{\mathrm{I}}_1}{\left(\mathrm{\mu } \mathrm{B}\right)}}}}{2\mathrm{\pi }\sqrt{{\displaystyle \frac{{\mathrm{I}}_1}{\left(\mathrm{\mu } \mathrm{B}\right)}}}}$

$T_2=\sqrt{2}T$