question_answer

Disregarding gravity, find the period of oscillation of the particle connected with four springs as shown in the figure.        (Given:\[\theta ={{45}^{o}},\beta ={{30}^{o}}\])           

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 \[T=2\pi \sqrt{\frac{m}{{{k}_{eq}}}}=2\pi \sqrt{\frac{m}{2k}}=\pi \sqrt{\frac{2m}{k}}\]