A) 0.09 m
B) 0.3 m
C) 0.03 m
D) 0.9 m
Correct Answer: B
Solution :
Angular velocity \[\omega =\sqrt{\left( \frac{k}{m} \right)}\]\[=\sqrt{\left( \frac{10}{10} \right)}=1\] Now \[u=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\]Þ\[{{y}^{2}}={{a}^{2}}-\frac{{{u}^{2}}}{{{\omega }^{2}}}\]\[={{(0.5)}^{2}}-\frac{{{(0.4)}^{2}}}{{{1}^{2}}}\] Þ \[{{y}^{2}}=0.9\] = \[y=0.3\,m\]
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