A)
\[{{T}^{2}}=T_{1}^{2}+T_{2}^{2}\]
B) \[{{T}^{-2}}=T_{1}^{-2}+T_{2}^{-2}\]
C) \[{{T}^{-1}}=T_{1}^{-1}+T_{2}^{-1}\]
D) \[T={{T}_{1}}+{{T}_{2}}\]
Correct Answer: B
Solution :
Key Idea: For a vertical spring-block system time period can be written as \[T=2\pi \sqrt{\frac{l}{g}}\]. We can write time period for a vertical spring-block system as \[T=2\pi \sqrt{\frac{l}{g}}\] Here, l is extension in the spring when the mass m is suspended from the spring. This can be seen as under: kl = mg (in equilibrium position) \[\Rightarrow \] \[\frac{m}{k}=\frac{l}{g}\] \[\therefore T=2\pi \sqrt{\frac{m}{k}}\] \[\therefore {{T}_{1}}=2\pi \sqrt{\frac{m}{{{k}_{1}}}}\Rightarrow \,\,{{k}_{1}}=4{{\pi }^{2}}\frac{m}{T_{1}^{2}}\,.....(i)\] \[{{T}_{2}}=2\pi \sqrt{\frac{m}{{{k}_{2}}}}\Rightarrow \,{{k}_{2}}=4{{\pi }^{2}}\frac{m}{T_{2}^{2}}....(ii)\] Since springs are in parallel, effective force constant \[k={{k}_{1}}+{{k}_{2}}\] \[\therefore T=2\pi \,\sqrt{\frac{m}{{{k}_{1}}+{{k}_{2}}}}\] \[\Rightarrow {{k}_{1}}+{{k}_{2}}=4{{\pi }^{2}}\frac{m}{{{T}^{2}}}...(iii)\] Substituting value of \[{{k}_{1}}\] and \[{{k}_{2}}\] from Eqs. (i) and (ii) in Eq. (iii) we get \[4{{\pi }^{2}}\frac{m}{T_{1}^{2}}+4{{\pi }^{2}}\frac{m}{T_{2}^{2}}=4{{\pi }^{2}}\frac{m}{{{T}^{2}}}\] \[\Rightarrow \frac{1}{{{T}^{2}}}=\frac{1}{T_{1}^{2}}+\frac{1}{T_{2}^{2}}\] \[\Rightarrow {{T}^{-2}}=T_{1}^{-2}+T_{2}^{-2}\]
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