Proving Simple Harmonic Motion (direction of acceleration) A particle of mass $m=0.5kg$, is attached to a spring of natural length $l=0.6m$ and modulus of elasticity $\lambda=60N$, and the setup is on a horizontal smooth table. The other end of the spring is attached to a fixed point $A$ on the table. The particle is then pulled so that the distance $AP=0.9m$, and is then released from rest. ($P$ is to the right of $A$, and the tension in the spring is $T$).
If you take $x$ (displacement from the centre of oscillation) as increasing to the right, then you can prove SHM if you can get equation that is like this:
$$a = -w^2$$
So
$$-T =ma$$ (The $T$ is negative because $x$ increases to the right making right positive, and $T$ is to the left, and $a$ is positive because in SHM $a$ is always in the direction of $x$ increasing.)
$$-\frac{\lambda x}{l} = ma$$
$$-\frac{60x}{0.6} = 0.5a$$
$$-200x = a$$ (which fits the equation at the top so its SHM.)
However if you take $x$ to increase to the left:
$$T = ma$$ ($T$ is positive because $x$ increases to the left making left positive and $T$ is to the left, $a$ is positive because, again, in SHM $a$ is always in the direction of $x$ increasing.)
$$\frac{\lambda x}{l} = ma$$
$$\frac{60x}{0.6} = 0.5a$$
$$200x = a$$ (which doesn't fit the equation because there is no negative.)
So I don't understand what I'm not getting right in the second part? how can changing the defining of the direction of $x$ increasing have such an effect?
(Instead of changing the direction of $x$ increasing I could have said the particle $P$ is pushed so that the spring is compressed and $AP=0.3m$ making the $T$ act towards the right which is the same direction as $x$ increasing).
 A: I think you are confused about which direction $T$ is acting in. You just have to look at the resulting forces and apply Newton's second law: $\sum{F}=ma$. To find the direction of the force and acceleration you just choose a direction, for instance to the right. This direction also has to apply to the position and velocity, since they are an integration of the acceleration. This is what you meant with: "$a$ is always in the direction of $x$ increasing"? You now just need to find an expression for the resulting force on the particle. A good way to check if you have the sign right is to make a free body diagram.
PS: the modulus of elasticity from your textbook has probably the unit of force per strain, but since unit of strain is distance per distance, the net unit it force. However I do find it weird that your textbook is calling it modulus of elasticity, since normally it has the unit of pressure. I myself prefer a spring constant in these kind systems, since they are more directly connected to the displacement.
