A) Motion remains SHM and period remains nearly constant
B) Motion remains SHM with time period = 2T
C) Motion remains SHM with time period = 4T
D) Motion remains SHM and time period = T/2
Correct Answer: B
Solution :
Time period of oscillation of magnet having mass m is given by \[T=2\pi \sqrt{\frac{I}{MB}}\,\,=\,\,2\pi \,\sqrt{\frac{m{{k}^{2}}}{MB}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,T\propto \sqrt{m}\] If \[{{T}_{1}}\], is time of oscillation when its mass is m and \[{{T}_{2}}\] is time of oscillation when its mass is 4m \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{m}{4\,m}}=\frac{1}{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{T}_{2}}=2{{T}_{1}}\] \[=\,\,\,2T\] Motion will remains SHM but time period is doubled.
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