A) \[T={{t}_{1}}+{{t}_{2}}\]
B) \[{{T}^{2}}=t_{1}^{2}+t_{2}^{2}\]
C) \[{{T}^{-1}}=t_{1}^{-1}+t_{2}^{-1}\]
D) \[{{T}^{-2}}\,=t_{1}^{-2}+t_{2}^{-2}\]
Correct Answer: B
Solution :
Time period of spring \[T=2\pi \sqrt{\left( \frac{m}{k} \right)}\] k, being the force constant of spring. For first spring\[{{t}_{1}}=2\pi \sqrt{\left( \frac{m}{{{k}_{1}}} \right)}\] ...(i) For second spring \[{{t}_{2}}=2\pi \sqrt{\left( \frac{m}{{{k}_{2}}} \right)}\] ...(ii) The effective force constant in their series combination of springs is \[k=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\] \[\therefore \]Time period of combination \[T=2\pi \sqrt{\left[ \frac{m({{k}_{1}}+{{k}_{2}})}{{{k}_{1}}{{k}_{2}}} \right]}\] \[\Rightarrow \]\[{{T}^{2}}=\frac{4{{\pi }^{2}}m({{k}_{1}}+{{k}_{2}})}{{{k}_{1}}{{k}_{2}}}\] ?..(iii) From Eqs. (i) and (ii), we obtain \[t_{1}^{2}+t_{2}^{2}=4{{\pi }^{2}}\left( \frac{m}{{{k}_{1}}}+\frac{m}{{{k}_{2}}} \right)\] \[\Rightarrow \]\[t_{1}^{2}+t_{2}^{2}=4{{\pi }^{2}}m\left( \frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}} \right)\] \[\Rightarrow \]\[t_{1}^{2}+t_{2}^{2}=\frac{4{{\pi }^{2}}m({{k}_{1}}+{{k}_{2}})}{{{k}_{1}}{{k}_{2}}}\] \[\therefore \] \[t_{1}^{2}+t_{2}^{2}={{T}^{2}}\] [from Eq. (iii)]
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