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Mechanical Vibrations Test 1

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Mechanical Vibrations Test 1
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  • Question 1
    1 / -0
    A vibratory system consists of a mass 15.7 kg, a spring of stiffness 1000 N/m and a dashpot damper. What is the critical damping of the system in Ns/m­­­­_________?
    Solution

    Concept:

    The critical system is when the damping ratio become 1.

    Natural frequency:

    ωn=km

    2ξωn = c/m

    Damping ratio:

    ξ=ccc=c2km=c2mωn

    Calculation:

    m = 15.7 kg, k = 1000 N/m

    For ξ = 1

    c=cc=2km=21000×15.7=250.60rad/s

  • Question 2
    1 / -0
    A spring mass damper system consists of mass of 25 kg and spring of stiffness 22 kN/m. If the damping is provided only 45 % of critical damping, then calculate the value of time period of vibration for this system_____ s
    Solution

    Explanation:

    Given:

    Mass (m) = 25 kg, k = 22,000 N/m, Damping provided = 45 %

    ξ=CCc=0.45

    Naturalfrequency(ωn)=km

    ωn=2200025

    ωn = 29.66 rad/s

    Damped frequency (ωd)

    ωd=ωn1ξ2

    ωd = 26.49 rad/s

    Time period = 2π/ω

    ∴ T = 0.237 sec 

  • Question 3
    1 / -0

    Given a machine that is connected to the ground whose law of motion is given below. What is the maximum force in N that can get transmitted to the ground?

    3x ¨+14x˙+72x=200cos(9t)

    Solution

    Concept:

    Motion of Forced damped vibration:

    mx ¨+cx˙+kx=Fcos(ωt) ------(1)

    ωn=Keqm

    Damping factor: ξ=ccc=c2km

    T=FTFun=1+(2ξωωn)2(1(ωωn)2)2+(2ξωωn)2 

    Calculation:

    Given:

    3x ¨+14x˙+72x=200cos(9t)

    On comparing the above equation with equation 1

    m = 3 kg, c = 14 Ns/m, K = 72 N/m, ω = 9 rad/s, F = 200 N

    ωn=Keqm=723=4.9 rad/s

    ξ=c2km=142×72×3=0.4763

    ωωn=94.9=1.837

    T=FTFun=1+(2ξωωn)2(1(ωωn)2)2+(2ξωωn)2 

    T=1+(2×0.4763×1.837)2(1(1.837)2)2+(2×0.4763×1.837)2 =2.0152.95=0.683

    FT=T×Fun=0.683×200=136.6 N

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