A) \[\omega ={{\left( \frac{k}{m}-\frac{{{b}^{2}}}{4{{m}^{2}}} \right)}^{1/2}}\]
B) \[\omega ={{\left( \frac{k}{m}-\frac{b}{4m} \right)}^{1/2}}\]
C) \[\omega ={{\left( \frac{k}{m}-\frac{{{b}^{2}}}{4m} \right)}^{1/2}}\]
D) \[\omega ={{\left( \frac{k}{m}-\frac{{{b}^{2}}}{4{{m}^{2}}} \right)}^{1/2}}\]
Correct Answer: A
Solution :
Displacement of damped oscillator is given by \[x={{x}_{m}}{{e}^{-bt/2m}}\sin (\omega 't+\phi )\] where \[\omega '=\]angular frequency of damped oscillator \[=\sqrt{\omega _{0}^{2}-{{(b/2m)}^{2}}}\] \[=\sqrt{\frac{k}{m}-\frac{{{b}^{2}}}{4{{m}^{2}}}}\]
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