A) \[T=2\pi \sqrt{\frac{M}{k}}\]
B) \[T=2\pi \sqrt{\frac{M}{2\,k}}\]
C) \[T=2\pi \sqrt{\frac{2\,M}{k}}\]
D) None of these
Correct Answer: B
Solution :
As spring 1 is compressed by a small distance x, then springs 2 and 3 are expended by \[{{\operatorname{X}}_{2}}={{X}_{3}} = x\,\,cos 45{}^\circ =\,\,\frac{x}{\sqrt{2}}\] \[\operatorname{F}=-\left( {{F}_{1}}+{{F}_{2}}+{{F}_{3}} \right)\] The net force acting on black will be Magnitude of force in vertical direction \[\operatorname{F} = -\left( {{F}_{1}} + {{F}_{2}} cos 45{}^\circ +{{F}_{3}} cos 45{}^\circ \right)\] \[=\,\,-\left( kx+\frac{k{{x}_{2}}}{\sqrt{2}}+\frac{k{{x}_{3}}}{\sqrt{2}} \right)=-2kx\] So acceleration, \[a=\frac{F}{M}=-\frac{2k}{m}x=-{{\omega }^{2}}x\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\omega =\sqrt{\frac{2k}{M}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,T=\frac{2\pi }{\omega }\,\,=\,\,2\pi \,\sqrt{\frac{M}{2k}}\]You need to login to perform this action.
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