question_answer

A system is shown in the figure. The time period for small oscillations of the two blocks will be-

A) \[2\pi \sqrt{\frac{3m}{k}}\]                    

B) \[2\pi \sqrt{\frac{3m}{4k}}\]

C) \[2\pi \sqrt{\frac{3m}{8k}}\]                  

D) \[2\pi \sqrt{\frac{3m}{2k}}\]

Correct Answer: B

Solution :

Both the spring are in series \[{{K}_{eq}}=\frac{k(2k)}{k+2k}\,\,=\,\frac{2k}{3}\] Time period \[T\,\,=\,\,2\pi \sqrt{\frac{\mu }{{{K}_{eq}}}}\] where \[\mu =\frac{{{m}_{1}}.{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] here \[\mu =\frac{m}{2}\] \[\therefore \,\,\,T=2\pi \sqrt{\frac{m}{2}.\frac{3}{2k}}\,\,=\,\,2\pi \sqrt{\frac{3m}{4k}}\] Alternative method: \[\therefore \text{ }m{{x}_{1}}=m{{x}_{2}}\text{ }\Rightarrow \text{ }{{x}_{1}}={{x}_{2}}\] force equation for first block; \[{{x}_{1}}={{x}_{2}}\,\,\,\Rightarrow \,\,\,\frac{{{d}^{2}}{{x}_{1}}}{d{{t}^{2}}}+\frac{4k}{3m}\,\,\,{{x}_{1}}=0\] Put \[{{x}_{1}}={{x}_{2}}\,\,\,\Rightarrow \,\,\,\frac{{{d}^{2}}{{x}_{1}}}{d{{t}^{2}}}+\frac{4k}{3m}\,\,\,{{x}_{1}}=0\]