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In an oscillations exercise there is a spring attached to another spring, attached to a block.

Long story short: I have to find the global $k$. In the solutions it says: "Because the springs are massless, they act similar to a rope under tension, and the same force $F$ is exerted by each spring."

I don't really understand how the springs being massless, is an argument for saying that they exert the same force (being the global force) upon the block.

Anyone care to elaborate?

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Hint: Draw a free body diagram (FBD) with forces on a small bit of the spring. Next apply Newton's 2nd law. –  Qmechanic May 20 '12 at 17:26
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2 Answers

up vote 2 down vote accepted

Massless springs only means that you do not have to calculate accelerations of the spring center of mass.

If the middle spring (the one between end spring and block) would have some mass, then the forces on its ends would not be equal, i.e.

$$\sum_i F = F_\text{right} - F_\text{left} = m_\text{spring} a $$

$$F_\text{right} = F_\text{left} + m_\text{spring} a$$

If however, $m_\text{spring} = 0$, then the force of the end spring ($F_\text{left}$) and the force acted on the block ($F_\text{right}$) are the same.

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Oh, ofcourse. Strange that I didn't realize this, because I did understand the exact same principle for rods when studying statics. Thanks! –  Edward Stumperd May 23 '12 at 12:00
    
Yes, this is exactly the same principle, and massless rope is similar too. –  Pygmalion May 23 '12 at 12:05
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The easiest way to explain this is that force is the flow of momentum, force tells you how much momentum is flowing from object A to object B. For this problem, the springs are causing momentum to flow between the two ends. Since the momentum is conserved, the total momentum is either transmitted or builds up on the springs. Since the springs are massless, they cannot carry any momentum themselves, so the momentum flows through to the other side without loss.

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