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 i.e., \[{{v}_{max}}=\omega a\]\[\Rightarrow \omega =\frac{{{v}_{max}}}{a}=\frac{16}{4}=4\] \ \[v=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\]\[\Rightarrow 8\sqrt{3}=4\sqrt{{{4}^{2}}-{{y}^{2}}}\] \[\Rightarrow 192=16\,(16-{{y}^{2}})\]\[\Rightarrow 12=16-{{y}^{2}}\]\[\Rightarrow y=2\,cm.\]
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