m = 100 kg s = 800 kN/m L F = 400 N ω = 3000 rpm = 314.2 rad/s (using 1 rev = 2π rad) x 2 = x 1 / 4 m = 100\textsf{ kg}\\
s = 800\textsf{ kN/m}L\\
F = 400\textsf{ N}\\
\omega = 3000\textsf{ rpm}= 314.2\textsf{ rad/s}\\
\textsf{(using 1 rev = 2π rad)}\\
x_2 = x_1/4 m = 100 kg s = 800 kN/m L F = 400 N ω = 3000 rpm = 314.2 rad/s (using 1 rev = 2 π rad) x 2 = x 1 /4
We know that angular speed or natural circular frequency of free vibrations;
ω n = s m = 800 × 1 0 3 100 = 89.44 rad/s \omega_n= \sqrt\dfrac sm= \sqrt\dfrac{800×10³}{100}= 89.44 \textsf{ rad/s} ω n = m s = 100 800 × 1 0 3 = 89.44 rad/s
For the amplitude;
log e x 1 x 2 = a × 2 π ( ω n ) 2 − a 2 \log_e \dfrac{x_1}{x_2} = a× \dfrac{2π}{\sqrt{(\omega_n)²-a²}} log e x 2 x 1 = a × ( ω n ) 2 − a 2 2 π
log e 4 = 2 π a ( 89.44 ) 2 − a 2 \log_e4 = \dfrac{2πa}{\sqrt{(89.44)²-a²}} log e 4 = ( 89.44 ) 2 − a 2 2 πa
a = 2236 l o g ( 2 ) ( 25 ( π 2 + l o g 4 ) a =\dfrac{ 2236 log(2)}{(25 \sqrt{(π^2 + log4)}} a = ( 25 ( π 2 + l o g 4 ) 2236 l o g ( 2 )
a = 19.27 a=19.27 a = 19.27
c = a × 2 m = 19.27 × 2 × 10 = 385.4 N/m/s c= a × 2m = 19.27 × 2 × 10= 385.4\textsf{ N/m/s} c = a × 2 m = 19.27 × 2 × 10 = 385.4 N/m/s
x o = F s = 400 800 × 1 0 3 = 0.0005 m x_o = \dfrac Fs = \dfrac{400}{800× 10³}= 0.0005m x o = s F = 800 × 1 0 3 400 = 0.0005 m
Amplitude of the forced vibration;
x max = x o c 2 ω 2 s 2 + [ 1 − ω 2 ω n 2 ] 2 x_{\textsf{max}} = \dfrac{x_o}{\sqrt{\dfrac{c²\omega²}{s²}+[1-\dfrac{\omega²}{\omega_n²}]²}} x max = s 2 c 2 ω 2 + [ 1 − ω n 2 ω 2 ] 2 x o
= 0.0005 385. 4 2 × 314. 2 2 ( 800 × 1 0 3 ) 2 + [ 1 − 314. 2 2 89.4 4 2 ] 2 = \dfrac{0.0005}{\sqrt{\dfrac{385.4²×314.2²}{(800×10³)²}+[1-\dfrac{314.2²}{89.44²}]²}} = ( 800 × 1 0 3 ) 2 385. 4 2 × 314. 2 2 + [ 1 − 89.4 4 2 314. 2 2 ] 2 0.0005
= 0.00004408 m = 0.0044 m m = 0.00004408m = 0.0044mm = 0.00004408 m = 0.0044 mm
[0.044mm; 35.2 N]
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