A) \[a-b\]
B) \[\frac{2a-b}{3}\]
C) \[\frac{2{{a}^{2}}}{3a-b}\]
D) None of these
Correct Answer: C
Solution :
[c] When particle starts from mean position, \[x=A\,\cos \,\omega t\] \[\therefore \,\,\,(A-a)=A\,\cos (\omega \times 1)\] Or \[(A-a)=A\,cos\,\omega \] ?..(i) \[Also\,\,\,A-(a+b)=A\,\cos (\omega \times 2)\] Or \[A-(a+b)=A\,\cos \,2\omega \] ??(ii) After simplifying above equations, we get \[A=\left[ \frac{2{{a}^{2}}}{3a-b} \right]\]You need to login to perform this action.
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