A) \[\sqrt{2{{k}_{2}}/{{k}_{1}}}\]
B) \[\sqrt{2{{k}_{1}}/{{k}_{2}}}\]
C) \[2{{k}_{1}}/{{k}_{2}}\]
D) \[2{{k}_{2}}/{{k}_{1}}\]
Correct Answer: A
Solution :
Given,\[{{m}_{2}}=2{{m}_{1}}\] For a loaded spring, the angular frequency \[\omega =\sqrt{k/m}\] \[\therefore \] \[\frac{{{\omega }_{1}}}{{{\omega }_{2}}}=\sqrt{\frac{{{k}_{1}}}{{{k}_{2}}}\times \frac{{{m}_{2}}}{{{m}_{1}}}}=\sqrt{\frac{{{k}_{1}}}{{{k}_{2}}}\times 2}\] For equal maximum velocity,\[{{A}_{1}}{{\omega }_{1}}={{A}_{2}}{{\omega }_{2}}\] we have, \[\frac{1}{8}={{e}^{-4.36}}\]\[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{{{\omega }_{2}}}{{{\omega }_{1}}}=\sqrt{\frac{{{k}_{2}}}{{{k}_{1}}}\times 2}=\sqrt{\frac{2{{k}_{2}}}{{{k}_{1}}}}\]
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