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?

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

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.

| cite | improve this answer | |
  • $\begingroup$ Oh, ofcourse. Strange that I didn't realize this, because I did understand the exact same principle for rods when studying statics. Thanks! $\endgroup$ – Edward Stumperd May 23 '12 at 12:00
  • $\begingroup$ Yes, this is exactly the same principle, and massless rope is similar too. $\endgroup$ – Pygmalion May 23 '12 at 12:05
  • $\begingroup$ I've stumbled across the same problem and it's driving me crazy: just a clarification.. with the first equation you apply the Newton's Laws to the spring, as if it was a single rigid object, such as a rod... I can't understand why two ends of a spring, not rigidly vinculated, cannot accelerate one respect to the other.. Can you help me? thanks a lot! $\endgroup$ – Surfer on the fall Apr 24 '16 at 19:43
  • $\begingroup$ One end of the spring can of course accelerate in respect to the other end. What counts for Newton's law is acceleration of the centre of mass. Most probably acceleration of the centre of mass would be average of accelerations of both ends. $\endgroup$ – Pygmalion Apr 26 '16 at 17:13

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.