m=0.25 kg
L=0.01568 m
u(0)=0, u′(0)=0.2 m/s, where u(t) is position of the mass at any time t.
k=Lmg=0.015680.25×9.8=156.25 N/m
Now we can write the equation of the system: mu′′+ku=0
0.25u′′+156.25u=0
u′′+625u=0
The characteristic equation: r2+625=0, r=±25i .
The general solution is u(t)=C1sin(25t)+C2cos(25t)
u(0)=0=C1×0+C2×1=C2 , C2=0
u′(0)=0.2=25C1 , C1=0.008
So, u(t)=0.008sin(25t) m.
Equilibrium position: u(t)=0, 0.008sin(25t)=0, sin(25t)=0, 25t=πn, n∈Z .
The mass first return to its equilibrium position when n=1 or t=π/25.
Answer:
a. u(t)=0.008sin(25t) m.
b. t=π/25 s.
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