The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to $α$ times its original magnitude, where $α$ equals___ $\times 10^{- 3}$

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answer is 729.

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Detailed Solution

$A = A_{0} e^{(\frac{- b t}{2 m})}$

$A_{0} \to$ max.amplitude

$0.9 A_{0} = A_{0} e^{(\frac{- b (5)}{2 m})} - - - - (1)$

$\begin{array}{l} A = A_{0} e^{\frac{- b (15)}{2 m}} \\ A = {A_{0}}^{[e^{\frac{- b (15)}{2 m}}]} - - - - (2) \end{array}$

$\begin{array}{l} A = A_{0} {(0.9)}^{3} \\ A = 0.729 A_{0} \\ A = α A_{0} \\ α = 0.729 \\ α = 729 \times 10^{- 3} \\ α = 729 \end{array}$

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