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B.Tech - 8MAE6(x)
A2305412321
A.S.E.T
Ques 1: A weight attached to a spring of stiffness 525 N/m has a viscous damπng device. When the weight is displaced and released, without damper the period of vibration is found to be 1.8secs, and the ratio of consecutive amplitudes is 4.2 to 1.0. Determine the amplitude and phase when the force F=2Cos3t acts on the system.
ANSWER:
Given: K = 525 N/m; τ = 1.8secs; x1 = 4.2; x2 = 1.0; F = F0sinωt = 2cos3t
So, F0 = 2N, ω = 3 rad/sec
X = (F0/K) / (√ [1 - r2] 2 + [2ζr]2)
ωn = 2π/ τ = 3.49rad/sec
δ = ln(4.2/1.0) = 1.435
ζ = δ / √ (4π2+ δ2) = 0.22
r = ω/ωn = 2/3.49 = 0.573
r2 = 0.328
X = (2/525) √ [1– 0.328] 2 + [4*0.484*0.328]
X = 5.3mm
Phase angle = tan-1 …show more content…
The slider of mass 2 Kgs within the machine has a reciprocating motion with a stroke of 0.08 m. The speed is 3000 rpm. Assuming the motion of the πston to be harmonic, determine :
Amplitude of vibration of the machine.
ANSWER:
Given, M = 75 Kgs; m = 2 Kgs, K = 11.76*105 N/m.
For vibrations due to rotating unbalance,
Amplitude of vibration= (MX / me) = {r2 / ( √[1- r2]2 + [2 ζ r]2)}
e = stroke/2 = 0.08/2 = 0.04 m
ω = 2π (3000) / 60 = 314 rad/sec.
ωn = √K/m = √(11.76*105 /75) = 125 rad/sec
ω / ωn = r = 314 /125 = 2.51
75 (X)/2(0.04) = (2.51)2 / {√(1-2.52)2 + (2*0.2*2.51)2}
Therefore,
X = 0.00125 m = 1.25 mm
Ques 4: The damped natural frequency of a system as obtained from a free vibration test is 9.8 cps. During a forced vibration test with a harmonic excitation on the same system, the frequency of vibration corresponding to peak amplitude was found to be 9.6 cps. Determine the damπng factor for the system and natural frequency.
ANSWER: ωd = 9.8 cps, ωp = 9.6 cps
(ωp / ωn) = (√1 -2ζ2)
ωn = ωd / (√1 -2ζ2)
Therefore,
[ {ωp * (√1 -2ζ2)} / ωd ] = (√1 -2ζ2)
Solving for ζ:
ζ = 0.196
ωn = ωd / (√1 -2ζ2) = 10
B.Tech - 8MAE6(x)
A2305412321
A.S.E.T
Ques 1: A weight attached to a spring of stiffness 525 N/m has a viscous damπng device. When the weight is displaced and released, without damper the period of vibration is found to be 1.8secs, and the ratio of consecutive amplitudes is 4.2 to 1.0. Determine the amplitude and phase when the force F=2Cos3t acts on the system.
ANSWER:
Given: K = 525 N/m; τ = 1.8secs; x1 = 4.2; x2 = 1.0; F = F0sinωt = 2cos3t
So, F0 = 2N, ω = 3 rad/sec
X = (F0/K) / (√ [1 - r2] 2 + [2ζr]2)
ωn = 2π/ τ = 3.49rad/sec
δ = ln(4.2/1.0) = 1.435
ζ = δ / √ (4π2+ δ2) = 0.22
r = ω/ωn = 2/3.49 = 0.573
r2 = 0.328
X = (2/525) √ [1– 0.328] 2 + [4*0.484*0.328]
X = 5.3mm
Phase angle = tan-1 …show more content…
The slider of mass 2 Kgs within the machine has a reciprocating motion with a stroke of 0.08 m. The speed is 3000 rpm. Assuming the motion of the πston to be harmonic, determine :
Amplitude of vibration of the machine.
ANSWER:
Given, M = 75 Kgs; m = 2 Kgs, K = 11.76*105 N/m.
For vibrations due to rotating unbalance,
Amplitude of vibration= (MX / me) = {r2 / ( √[1- r2]2 + [2 ζ r]2)}
e = stroke/2 = 0.08/2 = 0.04 m
ω = 2π (3000) / 60 = 314 rad/sec.
ωn = √K/m = √(11.76*105 /75) = 125 rad/sec
ω / ωn = r = 314 /125 = 2.51
75 (X)/2(0.04) = (2.51)2 / {√(1-2.52)2 + (2*0.2*2.51)2}
Therefore,
X = 0.00125 m = 1.25 mm
Ques 4: The damped natural frequency of a system as obtained from a free vibration test is 9.8 cps. During a forced vibration test with a harmonic excitation on the same system, the frequency of vibration corresponding to peak amplitude was found to be 9.6 cps. Determine the damπng factor for the system and natural frequency.
ANSWER: ωd = 9.8 cps, ωp = 9.6 cps
(ωp / ωn) = (√1 -2ζ2)
ωn = ωd / (√1 -2ζ2)
Therefore,
[ {ωp * (√1 -2ζ2)} / ωd ] = (√1 -2ζ2)
Solving for ζ:
ζ = 0.196
ωn = ωd / (√1 -2ζ2) = 10