Question

In damped oscillations, damping force is directly
proportional to speed of oscillator. If amplitude
becomes half of its maximum value in 1s, then after
2 s amplitude will be (Initial amplitude = A.)
​

Answer 1

Explanation:

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Answer 2

Given:

In damped oscillations, damping force is directly proportional to speed of oscillator. Amplitude becomes half of its maximum value in 1s.

To find:

Amplitude after 2 sec?

Calculation:

$\therefore \: A = A_{0} {e}^{ - \gamma t}$

At t = 1 sec ,

$\implies \: \dfrac{A_{0}}{2} = A_{0} {e}^{ - \gamma (1)}$

$\implies \: \dfrac{1}{2} = {e}^{ - \gamma}$

$\implies \: \dfrac{1}{2} = \dfrac{1}{ {e}^{ \gamma } }$

$\implies \: {e}^{ \gamma } = 2 \: \: \: \: . \: . \: . \: .(1)$

Now, when t = 2 sec:

$\therefore \: A = A_{0} {e}^{ - \gamma t}$

$\implies\: A = A_{0} {e}^{ - \gamma (2)}$

$\implies\: A = \dfrac{A_{0}}{ {e}^{2 \gamma } }$

$\implies\: A = \dfrac{A_{0}}{ {({e}^{\gamma })}^{2} }$

$\implies\: A = \dfrac{A_{0}}{{(2)}^{2} }$

$\implies\: A = \dfrac{A_{0}}{4}$

So, amplitude becomes ¼ th after 2 sec.