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Difficult but interesting standing waves question

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data
    A 160 g rope 4m long is fixed at one end and tied to a light string of same length to other end.Its tension is 400N.
    What is the wave length of first overtone?

    2. Relevant equations
    λ=v/f
    v=√(T/μ)
    μ=mass per unit length
    T=tension
    f=frequency
    3. The attempt at a solution
    Well,
    i know to find the wave length when the string is uniform density.Here i am unable to think how this works when we have two strings of different μ.Please guide me.First overtone is fundamental_3.JPG
     
  2. jcsd
  3. Oct 15, 2013 #2
    There are some conditions for these kind of problems, known as boundary conditions.

    1. Tension would be same in both the strings.
    2. Frequency would be same for both strings.

    when a heavy string is connected to a lighter string, that junction point is considered to be an antinode. I have given you pretty much every information you need to solve this question!!!
     
  4. Oct 15, 2013 #3
    Nihal are you from india.I am from india.
     
  5. Oct 15, 2013 #4
    Well, yes I am. But that is irrelevant to the question.:wink:
     
  6. Oct 15, 2013 #5
    are you preparing for iitjee??
     
  7. Oct 15, 2013 #6
    Why? Are you?
     
  8. Oct 15, 2013 #7
    Yes i am preparing for iitjee without coaching.
     
  9. Oct 15, 2013 #8
    good luck. If you want talk about iitjee and all that, use private message.
     
  10. Oct 16, 2013 #9
    Still not getting the problem
     
  11. Oct 16, 2013 #10
    What is it that you are not getting???
     
  12. Oct 16, 2013 #11
    I am not understanding how will the both strings work in first overtone. That is i am not getting the root concept.
     
  13. Oct 16, 2013 #12
    its not given that the second string is fixed at the other end. You just need to worry about the first string. If you are aware of boundary conditions, then you should not face any problem.
     
  14. Oct 16, 2013 #13
    What do we mean by light string??How will it interfere with the heavy string while making one node??
     
  15. Oct 16, 2013 #14
    light string means less linear mass density. The interference question is an interesting one, I'll suggest you read your textbook about this. It has to do with something with inertia.....let pulse travels through heavy string, when the pulse reaches the junction, two extreme conditions can be that the junction is a node or a antinode (intermediate conditions also happen). But when the pulse comes from the heavy string, it has more inertia compared to lighter string so it leads to an antinode in an extreme situation. The question you have given makes exactly this assumption to solve it.
     
  16. Oct 16, 2013 #15
  17. Oct 16, 2013 #16
    So that means that the light string has no effect on the heavy string,isn't it?
     
  18. Oct 16, 2013 #17
    Thanks for your precious guidance
     
  19. Oct 16, 2013 #18
    yes, you can say that.
    :thumbs:
     
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