A) \[\frac{\mu g}{4{{\pi }^{2}}{{V}^{2}}}\]
B) \[\frac{\mu g}{4{{\pi }^{2}}V}\]
C) \[\frac{\mu }{4{{\pi }^{2}}{{V}^{2}}g}\]
D) \[\frac{\mu }{2{{\pi }^{2}}{{V}^{2}}}\]
Correct Answer: A
Solution :
\[{{f}_{\max }}=\mu mg,\,{{a}_{\max }}=\mu g.\] If A is the amplitude \[{{a}_{\max }}=A{{\omega }^{2}}=4{{\pi }^{2}}A{{V}^{2}}=\mu g\] Therefore, \[A=\frac{\mu g}{4{{\pi }^{2}}{{V}^{2}}}.\]
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