A) \[{{t}_{0}}^{-1}={{t}_{1}}^{-1}+{{t}_{2}}^{-1}\]
B) \[{{t}_{0}}={{t}_{1}}+{{t}_{2}}\]
C) \[{{t}_{0}}^{2}={{t}_{1}}^{2}+{{t}_{2}}^{2}\]
D) \[{{t}_{0}}^{-2}={{t}_{1}}^{-2}+{{t}_{2}}^{-2}\]
Correct Answer: D
Solution :
\[t=2\pi \sqrt{\frac{m}{K}}\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,K=Const.\,{{t}^{-2}}\] Here the springs are joined in parallel. So \[{{K}_{0}}={{K}_{1}}+{{K}_{2}}\]where \[{{K}_{0}}\] is resultant force constant \[\therefore \,\,\,Const.\,t_{0}^{-2}=Const.t_{1}^{-2}+Const.t_{2}^{-2}\] or, \[t_{0}^{-2}=t_{1}^{-2}+t_{2}^{-2}\]
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