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Kinetic energy density in a string (derivative)

  1. Apr 25, 2017 #1
    1. The problem statement, all variables and given/known data
    In our physics course, we were studying one dimensional waves in a string. There, our teacher stated that the kinetic energy in a small piece of a string is [tex]dK=\frac{1}{2}μdx\frac{\partial y}{\partial t}^2[/tex] were μ is the linear density of the string, so he claimed that [tex]\frac{\text{d}K}{\text{d}x}=\frac{1}{2}μ\frac{\partial y}{\partial t}^2[/tex] which is impossible for me to understand, since K is a function of both x and t.

    3. The attempt at a solution
    Shouldn't we find the derivative of K=K(x,t) as [tex]\frac{\partial K}{\partial x}=\frac{\partial }{\partial x}(\frac{1}{2}μx)\frac{\partial y}{\partial t}^2+\frac{\partial }{\partial x}(\frac{\partial y}{\partial t}^2)\frac{1}{2}μx[/tex]
    What am i missing?
     
    Last edited: Apr 25, 2017
  2. jcsd
  3. Apr 25, 2017 #2
    I am going to assume that K is your symbol for kinetic energy (T is the more classical choice).

    The statement for dK is correct. This is a differential mass multiplied by the square of the velocity and divided by 2.

    The kinetic energy (K) is a single time depenndent quantity, not a function of position. To find K(t), you would need to integrate the expression for dK over the length of the string.

    Why does he even bother to write dK/dx? Does he really mean a total derivative, or perhaps a partial derivative?
     
  4. Apr 26, 2017 #3
    Thanks for replying,
    it makes sense that kinetic energy is only a function of time physically, however by looking the equation above since y=y(x,t), shouldn't ∂y/∂t be also a function of both position and time?
     
  5. Apr 26, 2017 #4
    With y = y(x,t), then partial y wrt t is a function of both x and t, so that much is correct.
     
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