a vertical spring mass system has a mass of 0.5 kg and an initial deflection of 0.2 cm. find the spring stiffness and the natural fequency of the system.
Answers
Answer:
Given:
Vertical spring system
Mass of the system, m = 0.5 kg
initial deflection of the spring, x = 0.2 cm
Formula Used:
Spring force, F=kxF=kx
where, k is the spring stiffness is constant and deflection of the spring.
\begin{gathered}\omega=\sqrt \frac{k}{m}\\f=\frac{1}{2\pi}\sqrt \frac{k}{m}\end{gathered}
ω=
m
k
f=
2π
1
m
k
Calculations:
For vertical spring system
\begin{gathered}kx=mg\\k=\frac{mg}{x}\\k=\frac{0.5\times 9.8}{0.002} = 2450 N/m\end{gathered}
kx=mg
k=
x
mg
k=
0.002
0.5×9.8
=2450N/m
\begin{gathered}f=\frac{1}{2\pi}\sqrt \frac{k}{m}\\f=\frac{1}{2\pi}\sqrt \frac{2450}{0.5} =11.15 Hz\end{gathered}
f=
2π
1
m
k
f=
2π
1
0.5
2450
=11.15Hz
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Answer:
75
Explanation:
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