A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm−1 and oscillates in a damping medium of damping constant 10−2 kg s−1.  The system dissipates its energy gradually.  The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to : Option 1)  2 s Option 2) 3.5 s Option 3)  5 s Option 4)  7 s

As we learnt in

Resultant amplitude in damped oscillation -

$A=A_{0}.e^{-\frac{bt}{2m}}$

$E=E_{0}.e^{-\frac{bt}{2m}}$

- wherein

$A= Amplitude$

$E= Energy$

Since the system dissipates its energy gradually and hence amplitude will also decrease with time

$a=a_{0}e^{-bt/m}$                            (i)

$\therefore$     Energy of vibration drop to half of its initial value $(\epsilon_{0})$

$E\ \alpha\ a^{2} \Rightarrow\ \; a\ \alpha \sqrt{E}$

$a=\frac{a_{0}}{\sqrt{2}}=\frac{bt}{m}=\frac{10^{-2}t}{0.1}=\frac{t}{10}$

$\therefore$    from equation(i)

$\frac{a_{0}}{\sqrt{2}}=a_{0}e^{-t/10}\ \; \Rightarrow\ \; \frac{1}{\sqrt{2}}=e^{-t/10}$       or    $\sqrt{2}=e^{t/10}$

$ln\sqrt{2}=\frac{t}{\sqrt{10}}\ \; \Rightarrow\ \; t = 3.5\ sec.$

Correct option is 2.

Option 1)

2 s

This is an incorrect option.

Option 2)

3.5 s

This is the correct option.

Option 3)

5 s

This is an incorrect option.

Option 4)

7 s

This is an incorrect option.

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