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Two bar magnets having same geometry with magnetic moments $M$ and $2 M$, are firstly placed in such a way that their similar poles are same side then its time period of oscillation is $T_{1}$. Now the polarity of one of the magnet is reversed then time period of oscillation is $T_{2},$ then
$T_{1} < T_{2}$
$T_{1}=T_{2}$
$T_{1}>T_{2}$
$T_{2}=\infty$
Solution
When similar poles are on same side time period of oscillation $T _{1}$ is given by $T _{2}=2 \pi \sqrt{\frac{ I _{1}+ I _{2}}{\left( m _{1}+ m _{2}\right) B _{ H }}}$
$m _{1}=2 m , m _{2}= m$
$T _{2}=2 \pi \sqrt{\frac{ I _{1}+ I _{2}}{(3 m ) B _{ H }}}$
When the polarity of magnet is reversed, time period of oscillation $T _{2}$ is given by
$T _{2}=2 \pi \sqrt{\frac{ I _{1}+ I _{2}}{\left( m _{1}- m _{2}\right) B _{ H }}}$
$T _{2}=2 \pi \sqrt{\frac{ I _{1}+ I _{2}}{ m B _{ H }}}$
From above Equations
$T _{1}< T _{2}$
