A) \[2\pi \sqrt{b}\]
B) \[\frac{2\pi }{b}\]
C) \[\frac{2\pi }{\sqrt{b}}\]
D) \[2\sqrt{\frac{\pi }{b}}\]
Correct Answer: C
Solution :
The relation between acceleration and displacement\[(x)\]for a body in SHM is \[a=-{{\omega }^{2}}x\] Given, \[a=-bx\] On comparing the two equations, we get \[{{\omega }^{2}}=b\] \[\therefore \] \[\omega =\sqrt{b}\] Since, \[\omega =\frac{2\pi }{T}\] \[\therefore \] \[\frac{2\pi }{T}=\sqrt{b}\Rightarrow T=\frac{2\pi }{\sqrt{b}}\]You need to login to perform this action.
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