# Direction of acceleration in Simple Harmonic Motion.

A particle of mass = 0.5kg, is attached to a spring of natural length = 0.6m and modulus of elasticity = 60N,and the setup is on a horizontal smooth table. The other end of the string is attached to a fixed point, A, on the table. The particle is then pulled so that the distance AP is 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)

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There's a difference between having a negative sign and being negative. If you're mass is towards the right, is your $x$ positive or negative? Should $T$ be positive or negative? – john Jun 9 '13 at 14:34