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Aar
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Homework Statement
An object with mass 0.222 kg is hung on a spring whose spring constant is 86.4 N/m. The object is subject to a resistive force given by -bv, where v is its velocity. The ratio of the damped frequency to the undamped (natural) frequency) is 0.886. If this system is subjected to a sinusoidal driving force given by
F(t)=(8.75 N)sin(0.901ωo t) ,
what is the (steady-state) amplitude (in cm) of the resulting sinusoidal motion
Homework Equations
A = A[itex]_{o}[/itex]/[itex]sqrt{( (1-r^2)^2 + (r^2/q^2) )}[/itex]
r = w[itex]_{d}[/itex]/w[itex]_{o}[/itex]
Q = (w[itex]_{o}[/itex]*M)/b
A[itex]_{0}[/itex] = Fm/k
F(t) = (Fm)cos(w[itex]_{d}[/itex]*t)
w[itex]_{d}[/itex] = [itex]\sqrt{w_{o}-(b/2m)^2}[/itex]
w^2 = k/m
The Attempt at a Solution
this is what I attempted initially and obtained an answer that was correct however in attempting other questions with different numbers the answers which I obtained were not correct. I am wondering if it was coincidental that the initial answer I obtained worked out to be a correct value or if it was the correct method and the program which I submitted the answer in was not registering the answers correctly.
initially solve for w[itex]_{0}[/itex] w[itex]_{d}[/itex] than obtain b
w[itex]_{0}[/itex] = [itex]\sqrt{86.4/0.222}[/itex] = 19.73
w[itex]_{d}[/itex] = w[itex]_{0}[/itex]*0.866 = 17.47
b = 2/m * [itex]\sqrt{(19.73)^2 - (17.47)^2}[/itex] = 4.0615
q = 19.73*0.222/4.0615 = 1.0783
A[itex]_{o}[/itex] = 8.75/86.4 = 0.10127
A = 0.10127/[itex]sqrt{( (1-0.866^2)^2 + (0.866^2/1.0783^2) )}[/itex] = 0.1192 m
convert to cm and the answer obtained is = 11.92cm
my uncertainty is wether or not the value in the sinusoidal driving force equation if the 0.901w[itex]_{o}[/itex] has any effect on determining w[itex]_{d}[/itex].
thanks for any help
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