The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10 s it will decrease to α times its original magnitude, where α equals :

Engineering

Physics

Basics of Simple Harmonic Motion

Damped Oscillation

Question

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10 s it will decrease to α times its original magnitude, where α equals :

0.81

0.729

0.7

0.6

A = A₀e^–rt

0.9A₀ = A₀e^{–r × 5}

αA₀ = A₀e^{–r × 15}

⇒α = (0.9)³ = 0.729

The amplitude of a damped oscillator decreases exponentially with time. For a damped oscillator, the amplitude at time t is given by:

$A = A_{0} e^{- b t}$

where $A_{0}$ is the original amplitude, $b$ is the damping constant, and $t$ is time.

Step 1: Find the damping constant

Given that after 5 seconds, amplitude becomes 0.9 times the original:

$0.9 A_{0} = A_{0} e^{- b \times 5}$

Divide both sides by $A_{0}$ :

$0.9 = e^{- 5 b}$

Take natural logarithm on both sides:

$\ln (0.9) = - 5 b$

So, $b = - \frac{\ln (0.9)}{5}$

Step 2: Find amplitude after another 10 seconds (total 15 seconds)

We need amplitude at $t = 15 s$ :

$A = A_{0} e^{- b \times 15}$

So, $α = \frac{A}{A_{0}} = e^{- 15 b}$

Step 3: Express in terms of known value

From Step 1, $e^{- 5 b} = 0.9$

Notice that $e^{- 15 b} = {(e^{- 5 b})}^{3} = {0.9}^{3}$

So, $α = {0.9}^{3} = 0.729$

Final Answer: α = 0.729

Formulae

Amplitude of damped oscillator: $A = A_{0} e^{- b t}$

Exponential decay property: $e^{- n b} = {(e^{- b})}^{n}$

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