A) Motion remains SH with time period = 4T
B) Motion remains SH and period remains nearly constant
C) Motion remains SH with time period = T/2
D) Motion remains SH with time period = 2T
Correct Answer: D
Solution :
If the magnet is displaced through an angle \[\theta \], the restoring torque in displaced position is \[\tau =-MH\,\,\sin \theta \] ...(i) Here, M = Magnetic moment of the magnet H \[\Rightarrow \] Horizontal component of earth's magnetic field but \[\tau =I\,\,\alpha \] and \[\sin \theta \approx 0\] for small angular displacement. Thus, Eq. (i) becomes \[I\,\alpha =-\,M\,H\,\theta \] ...(ii) or \[\alpha =-\frac{M\,H}{I}\,\theta \] \[=-\frac{M\,H}{(m{{\ell }^{2}}/12)}\theta \left( \because \,I=\frac{m{{\ell }^{2}}}{12} \right)\] \[=-k\theta \] where \[\text{k=}\frac{\text{M}\,\text{H}}{\text{m}{{\text{l}}^{\text{2}}}\text{/12}}\text{=a}\,\text{constant}\] If the mass of bar magnet is quadrupled, then k is again a constant. Hence, \[\text{a}\,\propto \,-\theta \] Thus, the motion is again simple harmonic. Now, from Eq. (ii) \[\left| \frac{\theta }{\alpha } \right|=\frac{I}{MH}\] The time period will be \[T=2\pi \,\sqrt{\left| \frac{\theta }{\alpha } \right|}=2\pi \,\sqrt{\frac{I}{MH}}\] or \[T\,\propto \,\,\sqrt{I}\] or \[T\propto \sqrt{m}\] \[\left( \because I=\frac{m{{l}^{2}}}{12} \right)\] \[\therefore \frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\] or \[\frac{T}{{{T}_{2}}}=\sqrt{\frac{m}{4m}}=\frac{1}{2}\] \[\therefore {{T}_{2}}=2T\]
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