Two bar magnets having same geometry with magnetic moments $m$ and $2 m$ , are firstly placed in such a way that their similar poles one are on same side. Then, its time period of oscillation is $T_{1}$ . Now, the polarity of the one of the magnet is reversed. Then, time period of oscillation is $T_{2}$ , then:

T_{1} < T_{2}

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T_{1} = T_{2}

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T_{1} > T_{2}

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T_{2} = \infty

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Solution

The correct option is A $T_{1} < T_{2}$
When similar poles are on same side time period of oscillation $T_{1}$ is given by
$T_{2} = 2 π \sqrt{\frac{I_{1} + I_{2}}{(m_{1} + m_{2}) B_{H}}} [g i v e n]$
$m_{1} = 2 m, m_{2} = m$
$T_{2} = 2 π \sqrt{\frac{I_{1} + I_{2}}{(3 m) B_{H}}} . . . . (i)$
When the polarity of magnet is reversed, time period of oscillation $T_{2}$ is given by
$T_{2} = 2 π \sqrt{\frac{I_{1} + I_{2}}{(m_{1} - m_{2}) B_{H}}}$
$T_{2} = 2 π \sqrt{\frac{I_{1} + I_{2}}{m B_{H}}} . . . . (i i)$
On comparing Eqs. (i) and (ii), we get
$T_{2} > T_{1}$
or $T_{1} < T_{2}$ .

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Two bar magnets having same geometry with magnetic moments m and 2m, are firstly placed in such a way that their similar poles one are on same side. Then, its time period of oscillation is T1. Now, the polarity of the one of the magnet is reversed. Then, time period of oscillation is T2, then: