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Question

(A) The time period of oscillation for a combination of two magnets in a magnetic field $B_{H}$ is given by $T = 2\pi \sqrt{\frac{I_{total}}{M_{total}

Vedclass · Accepted Answer

$T_{1} < T_{2}$

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(A) The time period of oscillation for a combination of two magnets in a magnetic field $B_{H}$ is given by $T = 2\pi \sqrt{\frac{I_{total}}{M_{total} B_{H}}}$.Case $1$: When similar poles are on the same side,the effective magnetic moment is $M_{eff} = M + 2M = 3M$. The moment of inertia is $I = I_{1} + I_{2}$.Thus,$T_{1} = 2\pi \sqrt{\frac{I_{1} + I_{2}}{3M B_{H}}}$.Case $2$: When the polarity of one magnet is reversed,the effective magnetic moment is $M_{eff} = |2M - M| = M$. The moment of inertia remains $I = I_{1} + I_{2}$.Thus,$T_{2} = 2\pi \sqrt{\frac{I_{1} + I_{2}}{M B_{H}}}$.Comparing $T_{1}$ and $T_{2}$,since the denominator in $T_{2}$ is smaller than in $T_{1}$,we have $T_{1} < T_{2}$.

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