Question 2
A spring-mass system, with a spring stiffness of 5,000 N/m, is subjected to a harmonic force of
magnitude 30 N and frequency 20 Hz. The mass is found to vibrate with an amplitude of 0.2 m.
Assuming that vibration starts from rest (x0 = ẋ0 = 0), determine the mass of the system. [10]
The static deflection
δst=F0k=30 N5000 M/m=0.006 mδ_{st} = \frac{F_0}{k} \\ = \frac{30 \;N}{5000 \;M/m} \\ = 0.006 \;mδst=kF0=5000M/m30N=0.006m
The frequency of the motion
ω=2πf=2π(20)=125.66 rad/sω= 2 \pi f \\ = 2 \pi (20) \\ = 125.66 \;rad/sω=2πf=2π(20)=125.66rad/s
The natural frequency of motion
X=δst(1−(ωωn)2)0.2=0.006(1−(125.66ωn)2)1−(125.66ωn)2=0.03125.66ωn=0.9848ωn=127.58 rad/sX = \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/sX=(1−(ωnω)2)δst0.2=(1−(ωn125.66)2)0.0061−(ωn125.66)2=0.03ωn125.66=0.9848ωn=127.58rad/s
The mass of the system:
ωn=km127.58=5000mm=0.3071 kgω_n = \sqrt{\frac{k}{m}} \\ 127.58 = \sqrt{\frac{5000}{m}} \\ m = 0.3071 \;kgωn=mk127.58=m5000m=0.3071kg
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