The static deflection
δ s t = F 0 k = 30 N 5000 M / m = 0.006 m δ_{st} = \frac{F_0}{k} \\
= \frac{30 \;N}{5000 \;M/m} \\
= 0.006 \;m δ s t = k F 0 = 5000 M / m 30 N = 0.006 m
The frequency of the motion
ω = 2 π f = 2 π ( 20 ) = 125.66 r a d / s ω= 2 \pi f \\
= 2 \pi (20) \\
= 125.66 \;rad/s ω = 2 π f = 2 π ( 20 ) = 125.66 r a d / s
The natural frequency of motion
X = δ s t ( 1 − ( ω ω n ) 2 ) 0.2 = 0.006 ( 1 − ( 125.66 ω n ) 2 ) 1 − ( 125.66 ω n ) 2 = 0.03 125.66 ω n = 0.9848 ω n = 127.58 r a d / s X = \frac{δ_{st}}{(1-(\frac{ω}{ω_n})^2)} \\
0.2 = \frac{0.006}{(1 -(\frac{125.66}{ω_n})^2)} \\
1 -(\frac{125.66}{ω_n})^2 = 0.03 \\
\frac{125.66}{ω_n} = 0.9848 \\
ω_n = 127.58 \;rad/s X = ( 1 − ( ω n ω ) 2 ) δ s t 0.2 = ( 1 − ( ω n 125.66 ) 2 ) 0.006 1 − ( ω n 125.66 ) 2 = 0.03 ω n 125.66 = 0.9848 ω n = 127.58 r a d / s
The mass of the system:
ω n = k m 127.58 = 5000 m m = 0.3071 k g ω_n = \sqrt{\frac{k}{m}} \\
127.58 = \sqrt{\frac{5000}{m}} \\
m = 0.3071 \;kg ω n = m k 127.58 = m 5000 m = 0.3071 k g
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