A) 7.0s
B) 14.2 s
C) 15.9s
D) 26.6 s
Correct Answer: A
Solution :
The given, \[K=90\,N/m\] \[b=0.04\,kg/s\] \[{{\log }_{e}}=2=0.693\] \[m=200\,g=0.2\,kg\] The (instantaneous) amplitude of damped oscillations is given by \[{{A}_{t}}=A{{e}^{-bt/2m}}\] Let \[{{t}_{1}}\] be the time taken for the amplitude to drop to half of its initial value, that is at \[t={{t}_{1}}\], \[{{A}_{t}}=A/2\] Thus, \[\frac{A}{2}=A{{e}^{-b{{t}_{1}}/2m}}\] \[{{e}^{b{{t}_{1}}/2m}}=2\] \[\frac{b{{t}_{1}}}{2m}={{\log }_{e}}2\] \[{{t}_{1}}=\frac{2m}{b}{{\log }_{e}}2\] \[=\frac{2\times 0.2}{0.04}{{\log }_{e}}2=\frac{0.4}{0.04}\times 0.693\] \[=10\times 0.693=6.93\,s\] or \[\cong \] 7.0 s
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