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Coupled Oscillations

Setting the scene:

We have two particles of masses mm and 2m2m attached to the points of trisection of a light elastic string of natural length 3a3a and modulus of elasticity λ\lambda.

The string is then stretched between two fixed points on a horizontal surface a distance 6a6a apart.

So picturing this:

img2.png

Question: Write down the equations of motion for small longitudinal displacements from the equilibrium position.

So by newton's second law with the positive direction to the right:

mx¨=T2−T1m\ddot{x} = T_{2}-T_{1} and 2my¨=T3−T22m\ddot{y} = T_{3}-T_{2}

Now Hooke's law says: T=λll0T = \lambda\frac{l}{l_{0}} where TT is tension, λ\lambda is the modulus of elasticity, ll is the extension and l0l_{0} is the natural length.

So how would you work out T1T_{1}, T2T_{2} and T3T_{3}?
(edited 10 years ago)
Original post by miketree


Now Hooke's law says: T=λll0T = \lambda\frac{l}{l_{0}} where TT is tension, λ\lambda is the modulus of elasticity, ll is the extension and l0l_{0} is the natural length.

So how would you work out T1T_{1}, T2T_{2} and T3T_{3}?


You've really answered your own question - Hooke's Law.

Treat each part of the elastic string separately - since the masses are attached.

The modulus of elasticity is the same as that of the whole string.

The initial length for each part will be a third the length of the original string.

...
Reply 2
Avatar for miketree
miketree
OP
Original post by ghostwalker
You've really answered your own question - Hooke's Law.

Treat each part of the elastic string separately - since the masses are attached.

The modulus of elasticity is the same as that of the whole string.

The initial length for each part will be a third the length of the original string.

...


Right so we have T1=λxaT_{1} = \lambda\frac{x}{a}, I have no idea what to do with T2T_{2} and T3T_{3} though
(edited 10 years ago)
Original post by miketree
Right so we have T1=λxaT_{1} = \lambda\frac{x}{a}, I have no idea what to do with T2T_{2} and T3T_{3} though


Careful. The string has been stretched to twice its original length initially. So, the extension is going to be "a+x", not just x.

For T2, before being displaced the extension is going to be "a". Then it's reduced by "x" and increased by "y", hence the new extension is....
Reply 4
Avatar for miketree
miketree
OP
Original post by ghostwalker
Careful. The string has been stretched to twice its original length initially. So, the extension is going to be "a+x", not just x.

For T2, before being displaced the extension is going to be "a". Then it's reduced by "x" and increased by "y", hence the new extension is....


Hm ok that makes sense, so T2=λ(a−x+ya)T_{2} = \lambda\left(\frac{a - x + y}{a}\right) and T3=λ(a−ya)T_{3} = \lambda\left(\frac{a-y}{a}\right)?
(edited 10 years ago)
Original post by miketree
Hm ok that makes sense, so T2=λ(a−x+ya)T_{2} = \lambda\left(\frac{a - x + y}{a}\right) and T3=λ(a−ya)T_{3} = \lambda\left(\frac{a-y}{a}\right)?


Yep.

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