In the under-damped vibrating system, the amplitude of vibration-

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  1. Decreases exponentially with time
  2. Increases linearly with time
  3. Increases exponentially with time
  4. Decreases linearly with time

Answer (Detailed Solution Below)

Option 1 : Decreases exponentially with time
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Explanation:

If the system is underdamped, it will swing back and forth with decreasing size of the swing until it comes to a stop. Its amplitude will decrease exponentially.

\(x\left( t \right) = {e^{ - \xi {\omega _n}t}}\left( {A{e^{i{\omega _d}t}} + B{e^{ - i{\omega _d}t}}} \right)\)

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Important Points 

F2 S.S-D.K 05.09.2019 D1

Overdamped System: ζ  > 1

\(x(t) = A{e^{( - \xi + \sqrt {{\xi ^2} - 1} ){ω _n}t}} + B{e^{( - \xi - \sqrt {{\xi ^2} - 1} ){ω _n}t}}\)

This is the equation of aperiodic motion i.e. the system cannot vibrate due to over-damping. The magnitude of the resultant displacement approaches zero with time.

Underdamped: ζ < 1

\(x\left( t \right) = {e^{ - \xi {ω _n}t}}\left( {A{e^{i{ω _d}t}} + B{e^{ - i{ω _d}t}}} \right)\)

\(x(t) = A{e^{ - \xi {ω _n}t}}\sin ({ω _d} + \phi )\)

This resultant motion is oscillatory with decreasing amplitudes having a frequency of ωd. Ultimately, the motion dies down with time.

Critical Damping: ζ = 1

\(x(t) = (A + Bt){e^{ - {\omega _n}t}}\)

The displacement will be approaching to zero with shortest possible time.

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