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 102 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

     

 

Answers (1)

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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