A) \[{{T}^{-1}}={{t}_{1}}^{-1}+{{t}_{2}}^{-1}\]
B) \[{{T}^{2}}=t_{1}^{2}+t_{2}^{2}\]
C) \[T={{t}_{1}}+{{t}_{2}}\]
D) \[{{T}^{-2}}=t_{1}^{-2}+t_{2}^{-2}\]
Correct Answer: B
Solution :
[b] \[{{t}_{1}}=2\pi \sqrt{\frac{m}{{{k}_{1}}},}\,\,\,{{t}_{2}}=2\pi \sqrt{\frac{m}{{{k}_{2}}}}\] when springs are in series then, \[{{k}_{eff}}=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\] \[\therefore \,\,\,T=2\pi \sqrt{\frac{m({{k}_{1}}+{{k}_{2}})}{{{k}_{1}}{{k}_{2}}}}\] \[\therefore \,\,\,T=2\pi \sqrt{\frac{m}{{{k}_{2}}}+\frac{m}{{{k}_{1}}}}=2\pi \sqrt{\frac{t_{2}^{2}}{{{(2\pi )}^{2}}}+\frac{t_{1}^{2}}{{{(2\pi )}^{2}}}}\] \[\Rightarrow \,{{T}^{2}}=t_{1}^{2}+t_{2}^{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
