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 :
| [d] 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 \[k=\frac{M\,H}{m{{l}^{2}}/12}=\text{a}\,\text{constant}\] |
| If the mass of bar magnet is quadrupled, then k is again a constant. Hence, \[a\,\propto \,-\theta \]. |
| 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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