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In the double-spring horizontal mass spring system, can I still apply the SHM equations $a=-\omega^2x$ and $T=2π\left(\frac{m}{k}\right)^{1/2}$?

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In equilibrium assume that both springs are in a state of tension.
The two springs exert equal magnitude but opposite direction forces on the mass; the left hand spring exerts a force to the left and the right hand spring exerts a force to the right.

When the mass moves to the left away from the equilibrium position the left hand spring exerts a smaller force to the left and the right hand spring exerts a larger force to the right.
Now a smaller force to the left is equivalent to a larger force to the right so the net effect is that magnitude of the force on the mass due to the two springs is twice that due to one spring and the direction is to the right.

So what is the effective spring constant of the two springs in this arrangement as compared with the spring constant of one spring $k$?

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  • $\begingroup$ Although I am not down voting it your second paragraph is quite confusing to me . I mean I think I knew the answer but now I am getting confused. Can you explain para 2 again? $\endgroup$ – Utkarsh futous Apr 6 '17 at 14:44
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    $\begingroup$ @Utkarshfutous Moving the mass to the left reduces the length of the left spring, which means that its extension is less and so it exerts a smaller force to the left on the mass. A similar argument for the right spring which is extended more means that it exerts a larger force to the right on the mass. So you have a reduction of the force to the left and an increase in force to the right. So the two springs are working in tandem and the net force to the right on the mass due to two springs is twice the force exerted by one string. $\endgroup$ – Farcher Apr 6 '17 at 14:52
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A spring constant tells you the cost you pay to compress or elongate the spring. If a spring has a spring constant $k$, it means you need $kx$ force to hold a spring which has been compressed or elongated by $x$ units in place.

Note how you pay the same price for elongation and compression.

In the config you provided, for a horizontal motion $x$ towards right or left, one spring elongates by $x$ and the other contracts by $x$. So, you need a total force of $k_1x+k_2x$ to hold an $x$ displacement in place.

So instead of $$F=-kx$$ you have $$F=-(k_1+k_2)x$$.

This is essentially the same as the original equation, but with a different $k$. Therefore, all the formulae should work the same, but with the new effective spring constant.

Think about what would happen if both springs were on the same side of the box, connected one after another, rather than one on each side the way it is here. If you can answer this one, you got it.

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