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Question

A single degree of freedom system having mass 1 kg and stiffness 10 kN/m initially at rest is subjected to an impulse force of magnitude 5 kN for 10-4 seconds. The amplitude in mm of the resulting free vibration is

The correct answer is

5.0

Impulse is 5000×10⁻⁴=0.5 N s, giving initial velocity 0.5 m/s. Natural angular frequency is √(10000/1)=100 rad/s. Free-vibration amplitude is v₀/ωn=0.005 m=5 mm.

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Important Questions from Transmissibility and Magnification Factor

  1. When ω/ωn = 1, the transmissibility is

    where ω = frequency of excitation, rad/s

    ωn = Natural frequency, rad/s

  2. The Transmissibility of a vibrating system, for the values of (ω/ωn) >√2, will be

  3. A vibrating system having a mass of 30 kg is supported by 4 springs, each having a stiffness of k. The system runs at 420 rpm. If only 10% of the shaking force is allowed to be transmitted to the supporting structure, then the value of stiffness k will be close to:

  4. For a single degree of freedom viscous damped system, if the frequency ratio is greater than √2, it implies that:

  5. Dynamic Magnification Factor is computed as:

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