A) \[\pi \sqrt{\frac{2m}{k}}\]
B) \[\sqrt{\frac{2m\pi }{k}}\]
C) \[\sqrt{\frac{m\pi }{2k}}\]
D) \[\pi \sqrt{\frac{m}{2k}}\]
Correct Answer: A
Solution :
[a] \[2k\,{{\sin }^{2}}\theta ={{k}_{1}}\,or\,{{k}_{1}}=2k\,{{\sin }^{2}}\theta \] and \[{{k}_{2}}=2(2k){{\sin }^{2}}\beta \] Then\[{{k}_{eq}}={{k}_{1}}+{{k}_{2}}=2k[{{\sin }^{2}}\theta +2{{\sin }^{2}}\beta ]\] \[=2k[{{\sin }^{2}}{{45}^{o}}+2{{\sin }^{2}}{{30}^{o}}]=2k\left( \frac{1}{2}+\frac{1}{2} \right)=2k\] Then \[T=2\pi \sqrt{\frac{m}{{{k}_{eq}}}}=2\pi \sqrt{\frac{m}{2k}}=\pi \sqrt{\frac{2m}{k}}\]You need to login to perform this action.
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