# SHM spring system, Tension direction confusion.

Say a spring is on a horizontal table, and the particle is initially at the midpoint of the spring in equilibrium, and then the particle is displaced to one side.

I have attached a diagram, let this side be more to A.

Then I am confused as to which direction the tension in both sides of the spring will act and why.

My thoughts would be along the lines of:
- displaced more towards A, this side of the spring is compressed, the tension will therefore act in the direction AB acting to re-extend the spring (restoring)
- and vice versa with spring B, which is extended and therefore will act to compress
- of course these thoughts are wrong as the tensions would both then be in the same direction.

Thanks alot.

Also upon derviation I am using the time when the particle is intially displaced and so would conclude that the acceleration direction is in the direction BA

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cepheid
Staff Emeritus
Gold Member
My thoughts would be along the lines of:
- displaced more towards A, this side of the spring is compressed, the tension will therefore act in the direction AB acting to re-extend the spring (restoring)
- and vice versa with spring B, which is extended and therefore will act to compress

Sounds alright to me.

- of course these thoughts are wrong as the tensions would both then be in the same direction.

What is wrong with that?

What is wrong with that?

(I have attached a diagram showing the correct diagram and equations.)

Not applying this correct has led me to derive my equation wrong:

- The question is that initially a string has a particle p attached to its midpoint and is stretched at 4a: AB=4a and natural length is 2a.
- The particle is then released from rest at C, where AC - 3a/2, establish the equation of motion.

My thoughts were that the part ‘A’ is compressed and so will naturally act to restore, acting in the opposite diagram than shown in the diagram, and B would act naturally to restore its extension also acting in this direction.

- Mg(a+x)/a - mg(a-x)/a - mn[dx/dt]= mx[d^2x/dt^2] as a pose to the solution attached.

(judging by the initial conditions to set up the system - system is moved to the right (hence acceleration this way) and resistance opposes this, tensions as above.)

Thanks alot for any assistance !

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