A) \[2\pi \sqrt{\frac{m}{{{K}_{1}}+{{K}_{2}}}}\]
B) \[2\pi \sqrt{\frac{m\sin \theta }{{{K}_{1}}+{{K}_{2}}}}\]
C) \[2\pi \sqrt{\frac{m({{K}_{1}}+{{K}_{2}})}{{{K}_{1}}{{K}_{2}}}}\]
D) \[2\pi \sqrt{\frac{m\sin \theta ({{K}_{1}}+{{K}_{2}})}{{{K}_{1}}{{K}_{2}}}}\]
Correct Answer: A
Solution :
Let elongation of spring in equilibrium position is \[{{x}_{0}},\] then\[({{K}_{1}}+{{K}_{2}}){{x}_{0}}=mg\sin \theta \] So,\[T=2\pi \sqrt{\frac{{{x}_{0}}}{g\sin \theta }}=2\pi \sqrt{\frac{m}{{{K}_{1}}+{{K}_{2}}}}\]
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