A) \[\frac{2\pi }{K}\]
B) \[2\pi K\]
C) \[\frac{2\pi }{\sqrt{K}}\]
D) \[2\pi \sqrt{K}\]
Correct Answer: C
Solution :
On comparing with standard equation \[\frac{{{d}^{2}}y}{d{{t}^{2}}}+{{\omega }^{2}}y=0\] we get \[{{\omega }^{2}}=K\Rightarrow \omega =\frac{2\pi }{T}=\sqrt{K}\Rightarrow T=\frac{2\pi }{\sqrt{K}}\].You need to login to perform this action.
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