A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to $\frac{1}{1 0 0 0}$ of the original amplitude is close to :-

Question

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Answer 1

$A = A_{0} e^{- γ t}$
$A = \frac{A _{0}}{2}$ after 10 oscillations
$∵$ After 2 seconds
$\frac{A _{0}}{2} = A_{0} e^{- γ (2)}$
$2 = e^{2 γ}$
$ℓ n 2 = 2 γ$
$γ = \frac{ℓ n 2}{2}$
$∵ A = A_{0} e^{- γ t}$
$ℓ n \frac{A _{0}}{A} = γ t$
$ℓ n 1000 = \frac{ℓ n 2}{2} t$
$2 (\frac{3 ℓ n 1 0}{ℓ n 2}) = t$
$\frac{6 ℓ n 1 0}{ℓ n 2} = 1$
$t = 19.931 s e c$
$t \approx 20 s e c$

Answer 2

A = A_{0} e^{- γ t}

A = \frac{A _{0}}{2}

after 10 oscillations

∵

After 2 seconds

\frac{A _{0}}{2} = A_{0} e^{- γ (2)}

2 = e^{2 γ}

ℓ n 2 = 2 γ

γ = \frac{ℓ n 2}{2}

∵ A = A_{0} e^{- γ t}

ℓ n \frac{A _{0}}{A} = γ t

ℓ n 1000 = \frac{ℓ n 2}{2} t

2 (\frac{3 ℓ n 1 0}{ℓ n 2}) = t

\frac{6 ℓ n 1 0}{ℓ n 2} = 1

t = 19.931 s e c

t \approx 20 s e c