1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy in standing waves

Tags:
  1. May 10, 2016 #1

    Titan97

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    The ends of a stretched wire of length L are fixed at x=0 and x=L. In one experiment, the displacement of the wire is given by ##y=A\sin\left(\frac{\pi x}{L}\right)\sin(\omega t)## and its energy is ##E_1##. In another experiment, the displacement of wire is given by ##y=A\sin\left(\frac{2\pi x}{L}\right)\sin(2\omega t)## and its energy is ##E_2##. If ##E_2=kE_1##, find ##k## (k is a positive integer).

    2. Relevant equations
    None

    3. The attempt at a solution
    $$y=A\sin\left(\frac{\pi x}{L}\right)\sin(\omega t)=\frac{A}{2}\left[\cos\left(\frac{\pi x}{L}-\omega t\right)-\cos\left(\frac{\pi x}{L}+\omega t\right)\right]$$

    The given wave is formed by two waves travelling in opposite direction with amplitudes ##\frac{A}{2}##. At ##t=0##, the two waves completely cancel out each other. Hence the energy becomes zero. Isnt ##E_1## time dependent?
     
  2. jcsd
  3. May 10, 2016 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The waves only cancel each other in the displacement. This means the potential is zero at that moment. They do not cancel in the velocity. In fact, the string has its maximal velocity at that time.

    I suggest finding an expression for the kinetic energy of the string and compute it at t=0 when the potential is zero.

    No, the energies are not time dependent.
     
  4. May 10, 2016 #3

    Titan97

    User Avatar
    Gold Member

    Using ##\text{d}K=\frac{1}{2}(\mu \text{d}x) v_y^2##
    $$K=\frac{1}{4}\mu\omega^2A^2L\cos^2\omega t$$

    at ##t=0##
    $$E_1=\frac{1}{4}\mu\omega^2A^2L$$
    $$E_2=\frac{1}{4}\cdot 2\mu\cdot 4\omega^2A^2\frac{L}{2}=4E_1$$
     
  5. May 10, 2016 #4

    Titan97

    User Avatar
    Gold Member

    @Orodruin how can I compute potential energy as a function of time?
     
  6. May 11, 2016 #5

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You have to compare the length of the string to the rest length. The potential energy is the tension multiplied by this change in length.

    To get energy conservation with the waves you will need to use the first order approximation, but that was also used for deriving the wave equation so it is consistent.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Energy in standing waves
  1. Standing Wave (Replies: 2)

  2. Standing waves (Replies: 3)

  3. Standing waves (Replies: 2)

Loading...