A) \[2\sqrt{3\,}cm\]
B) \[\sqrt{3\,}cm\]
C) 1cm
D) 2cm
Correct Answer: D
Solution :
[d] At mean position velocity is maximum i.e., \[{{v}_{\max }}=\omega a\Rightarrow \omega \frac{{{v}_{\max }}}{a}=\frac{16}{4}=4\] \[\therefore \,\,v=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\Rightarrow 8\sqrt{3}=4\sqrt{{{4}^{2}}-{{y}^{2}}}\] \[\Rightarrow \,192=16(16-{{y}^{2}})\Rightarrow 12=16-{{y}^{2}}\] \[\Rightarrow \,y=2cm.\]
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