2.003SC Recitation 10 Notes: Natural Frequency From Deflection & Frequency Response Obtaining Natural Frequency from Spring Deﬂection Consider a spring whose unloaded length is as shown. Spring deflection when mass is placed on spring Unloaded spring length When a mass is placed on the spring, in the presence of gravity, the spring deﬂects due to the mass’s weight. The amount of the deﬂection can be seen to be δ= mg k k= mg δ so Recalling that the natural frequency is given by � ωn = k = m � or � ωn = g δ mg δm Steady State Frequency Response Consider a spring-mass-dashpot system subjected to a periodic forcing function. The equation of motion is mẍ + bẋ + kx = F0 cos(ωt) ẍ + b k 1 1 k · F0 cos(ωt) ẋ + x = F0 cos(ωt) = m m m m k ẍ + 2ζωn ẋ + ωn 2 x = ωn2 F0 cos(ωt) k Let F0 cos(ωt) = X static eiωt k x(t) = X out eiωt Then we can deﬁne the transfer function, H(jω) , as the ratio of the output to the input, H(jω) = 1 X out = ω 2 X static 1 − ( ωn ) + i2ζ ωωn The transfer function (a complex number) can be resolved into its magnitude and phase. The magnitude and phase are functions of the forcing frequency and are given by |H(ω)| = � 1 [1 − ( ωωn )2 ]2 + [2ζ( ωωn )]2 and � φ = tan−1 2ζ( ωωn ) 1 − ( ωωn )2 � Steady State Frequency Response - Problem Statement A large blower is mounted on a steel frame which is rigidly connected to the ﬂoor of a building. The blower’s rotor is unbalanced, and subjects the blower to a sinusoidally varying vertical load. © source unknown. All rights reserved. The blower weighs 500 lb. When it is placed on the frame, the frame deﬂects a vertical distance δ = 0.026 in. When the blower is turned on, the blower’s rotor spins at ω = 1750 rpm. When the blower is installed and turned on, it vibrates violently in the vertical direction. Your assistant has designed some braces which (s)he says will increase the vertical stiﬀness of the base by about a factor of 2, and believes this will reduce the vibration by about the same factor. Do you accept the proposal? Tasks • Draw a (lumped-parameter) model of the system. • Determine the system’s undamped natural frequency, ωn , and frequency ratio, • Estimate the magnitude of the system’s frequency response, |H(ω)| . • Determine the eﬀect of the proposed change. ω ωn . Steady State Frequency Response - Solution Model Frequencies The forcing frequency is ω = 1750 rev min · 1 60 min sec · 2π radians rev = 183.26 radians sec The deﬂection in SI units is, δ = 0.026in · 1 39.37 meters in = 0.00066 meters The undamped natural frequency (before the proposed change) is, (ωn )bef ore = k m = g δ = 9.81 0.00066 = 121.88 radians sec The frequency ratio (before the proposed change) is ( ωωn )bef ore = 183.26 121.88 = 1.504 Magnitude The magnitude of the frequency response is given by xstatic · |H(ω)| = F0 · k 1 [1 − ( ωωn )2 ]2 + [2ζ( ωωn )]2 Evaluating, xstatic · |H(ω)|bef ore = F0 · (0.8) k After the proposed change, Frequencies The undamped natural frequency is, √ (ωn )af ter = 2(ωn )bef ore = 172.36 radians sec The frequency ratio is ( ωωn )af ter = 183.26 172.36 = 1.063 Magnitude After the proposed change, the magnitude of the vibration is, xstatic · |H(ω)|af ter = F0 · (7.667) k Eﬀect of the change... 7.667 |H(ω)|af ter = = 9.58 bef ore |H(ω)| 0.8 The vibration will INCREASE by a factor of about 10 !!! DON’T MAKE THE CHANGE!!!