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