solution
comparing the standard equation
x (t) = x0 cm Cos(ωt)
A)so force conatant
k=mω2=2.25×(3.25)2=23.75N/m
B)time
t=ω2π=3.252×3.14=1.93sec
C)velocity is given
v=dtdx=−5.50×3.25sin(3.25t)
here maximum velocity is v=0.1787 cm/sec
D)force is given
F=mdt2d2x=−2.25×5.50×3.25×3.25cos(3.25t)
so maximum force
F=2.25×5.50×3.25×3.25=130.71N
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