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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).

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to the person who suggestedthe edit. my textbook says modulus of elasticity and that its units are newtons –  Jonathan. Jun 9 '13 at 14:12
    
You can ping anyone who edited . Also the units for Young's modulus is same as pressure. –  ABC Jun 9 '13 at 14:48

1 Answer 1

I think you are confused about which direction $T$ is acting on. You just have to look at the resulting forces and apply Newton's second law: $\sum{F}=ma$. And 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 an 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 self prefer a spring constant in these kind systems, since they are more directly connected to the displacement.

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thanks for your answer, I think this diagram might explain my question better, i.stack.imgur.com/MlKkD.jpg my problem is with (2), it doesn't have a minus sign. –  Jonathan. Jun 9 '13 at 16:05
    
You do not need 2 expressions to describe the acceleration, since if the displacement is negative than the spring is compressed and therefore the tension is negative. –  fibonatic Jun 9 '13 at 19:33

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