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Nodes and anti nodes

  1. May 3, 2014 #1
    a standing wave of time period T is set up in a string clamped between two rigid supports. at t=0 antinode is at its maximum diaplacement 2A. at what time will the energy of antinode be equal to that of a node?


    what i know is...
    that node has zero amplitude..n antinode has maximum amplitude. and that the maximum potential energy= 1/2*ρ*ω2A2. which also equals the kinetic energy..
    moreover it can be seen that the amplitude of the given wave is 2A.
    T=2∏/ω.
    but i am unable to relate the above information in order to solve the given query.
    please tell me how can i find the energy of an antinode and a node?

    i am very weak in physics... please guide me through it.
     
  2. jcsd
  3. May 3, 2014 #2

    haruspex

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    I assume by 'energy' of a point on the string the question means the sum of KE and PE at that point.
    What form does the PE take in a vibrating string? How will that be distributed along the string?
     
  4. May 4, 2014 #3
    energy of a point on a string= 1/2Δmvy2 + 1/2Δmω2y2 (where y is the amplitude of that point)
    so this means at a node total energy woul become equal to 1/2Δmvy2 since y=0 and the total energy at the anti node would become equal to 1/2Δmvy2 + 1/2Δmω2(2A)2.

    am i right about this?

    and if so... how do i get to know the velocity at the node and antinode? and how do i relate this to the timepriod of the wave?

    is the velocity of an anti node zero?

    moreover value of ω is also not given..

    please please guide me more..
     
  5. May 4, 2014 #4
    The velocity at the nodes is zero since they don't move. The formula you provided for the energy looks right but it isn't.
     
  6. May 4, 2014 #5
    then how do i find the formula? i searched it all over.. and even thought as much as i could... but i am getting nowhere :cry:
     
  7. May 4, 2014 #6
    Your expression for the kinetic energy is correct but the expression for the potential energy isn't. The potential energy is given by the stretching of the spring, not by its displacement.

    it is proportional to the square of the derivative of the wave function (dy/dx)2
     
  8. May 4, 2014 #7
    sir... but the following link says exactly what i said: http://hyperphysics.phy-astr.gsu.edu/hbase/waves/powstr.html (please refer to the section: "energy in a string wave" of this link)
     
  9. May 4, 2014 #8
    That page is wrong, unfortunately. The final result for the total energy propagation is correct (for a traveling harmonic wave), but the details of the derivation are not.
     
  10. May 4, 2014 #9
    so according to you.. the potential energy should be equal to 1/2Δmω2(dy/dx)2 ??
     
  11. May 4, 2014 #10
    this means the total energy at the node is zero.. am i right?
     
  12. May 4, 2014 #11
    Almost, but not quite. It's given by

    ΔW = (1/2) Δm v2 (dy/dx)2

    where v is the wave speed.
     
  13. May 4, 2014 #12
    this means energy at a node is zero. am i right?
     
  14. May 4, 2014 #13
    No, the potential energy at the anti-node is always zero and the kinetic energy at the node is always zero. The energy flips back and forth between Kinetic and Potential as it bounces back and forth between the anti-node and the node.
     
  15. May 4, 2014 #14
    so... back to the asked question..
    in the query.. we have to find the time wen energy at node=energy at anti node.
    KE of node = PE of anti node
    ∴ i/2Δmv2 = 1/2ΔmV(dy/dx)2
    v=√T/μ
    would i be wrong if i relate this in the equation... since we must need to relate the timeperiod with energy?
     
  16. May 5, 2014 #15
    so... back to the asked question..
    in the query.. we have to find the time wen energy at node=energy at anti node.
    KE of node = PE of anti node
    ∴ 1/2Δmv2 = 1/2ΔmV(dy/dx)2
    v=λf and f=1/T
    ∴ v=λ/T
    would i be wrong if i relate this in the equation... since we must need to relate the timeperiod with energy?
    and my other doubt is... that equation of wave is not given... so how to find dy/dx and λ?
     
  17. May 5, 2014 #16
    The formula for a standing wave is just the product of two sinusoidal functions.

    y = A sin(kx + δ) sin(ωt + ø),

    Where δ and ø are just phases
     
  18. May 8, 2014 #17
    well... i seem to be going no where towards solving my question :(
     
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