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99wattr89
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I can't make this question work, so I'm hoping that someone here will be able to help guide me towards a solution.
I began with F=ma, and wrote down the equations of motion for each of the masses.
a) 2mx..1 = -kx1 -k(x1 -x2)
and
b) mx..2 = -kx2 +k(x1 -x2)Then I added b to a, and subtracted b from a to get;
c) m(2x..1 +x..2) = -k(x1 +x2)
and
d) m(2x..1 -x..2) = -3k(x1 -x2)Then I added d to c, and subtracted d from c to get;
e) mx..1 = -kx1
and
f) mx..2 = kx1 -2kx2I then tried to solve e, using the trial solution x1 = αcosωt + βsinωt
But I don't have boundary conditions, so I can't find α and β, just ω, which I found to be √(k/m)I then tried putting this expression into f to allow me to solve it for x2, but I realized that I could only solve it if the ω term associated with x1 was the same as the ω term associated with x2 - otherwise the sin and cos terms can't be canceled out. And if ω is the same for both then my method has to be wrong anyway, because I'm meant to find two values for ω, and as the question states, neither is √(k/m)Can anyone help me here?
Thank you.
I began with F=ma, and wrote down the equations of motion for each of the masses.
a) 2mx..1 = -kx1 -k(x1 -x2)
and
b) mx..2 = -kx2 +k(x1 -x2)Then I added b to a, and subtracted b from a to get;
c) m(2x..1 +x..2) = -k(x1 +x2)
and
d) m(2x..1 -x..2) = -3k(x1 -x2)Then I added d to c, and subtracted d from c to get;
e) mx..1 = -kx1
and
f) mx..2 = kx1 -2kx2I then tried to solve e, using the trial solution x1 = αcosωt + βsinωt
But I don't have boundary conditions, so I can't find α and β, just ω, which I found to be √(k/m)I then tried putting this expression into f to allow me to solve it for x2, but I realized that I could only solve it if the ω term associated with x1 was the same as the ω term associated with x2 - otherwise the sin and cos terms can't be canceled out. And if ω is the same for both then my method has to be wrong anyway, because I'm meant to find two values for ω, and as the question states, neither is √(k/m)Can anyone help me here?
Thank you.