Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a block of mass $m$ attached to a spring. Let it oscillate at a frequency $f$. Now each part of the spring is in SHM. so this means a wave is propagating through this spring.bCan this wave be reflected at the fixed end of the spring resulting in the formation of standing waves?

share|cite|improve this question
Yes, in valvetrains it the reflected wave in the valve spring that cause surge and loss of contact at high rpm. – ja72 May 19 '13 at 16:23

Well, the reflection of a wave at the end happens always. One can picture this by imagining the succesive atoms being pushed off the equilibrium position as the wave propagates. Since the endpoint is fixed, it has nowhere to be pushed but the few atoms near it (I am considering idealized linear chain for simplicity) that have already being perturbed will, after having passed through equilibrium again, pass into the opposite direction.

For transversal waves (as those you have on strings of a guitar) this means that the wave perturbation will change from "up" to "down" at the end (and vice versa) while for the longitudinal waves (as those in your spring) there is a change from "compressed" to "streched" (and vice versa).

enter image description here

share|cite|improve this answer
endpoint is fixed? I think that the endpoint is the last atom in the chain. IMO, the gray line is a graphical helper and the 'length' of a body is always changing. – Helder Velez Aug 17 '11 at 10:17
@Helder: you can imagine that there are two more dots at both ends of the line that are held fixed (but since they are fixed, they don't matter anyway...). In any case, the picture makes it clear that normal modes also exist for longitudinal waves which is what OP was after... – Marek Aug 17 '11 at 11:23
I upvoted the answer because it is clear. But there are no 'rigid-bodys' in the real world. We have atom's aggregated by EM and none of them is 'fixed' (There is no 'anchor' available). – Helder Velez Aug 17 '11 at 13:01
Great pictures. I might add that if you take the original mass on a spring, replace the spring with one whose stiffness is exactly half, put the mass in the middle of the spring instead of the end, and then fasten a string of these units together end to end, you get the case shown in the very first diagram above. You only get standing waves at that frequency; to get travelling waves, you have to go lower down in the pictures and drive the system at a much lower frequency it approximates a continuous medium. – Marty Green Aug 17 '11 at 13:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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