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My textbook in deriving the wave equation for a one dimensional elastic string stated that the horizontal direction force is constant.I understand that the horizontal components of the tensions on either side of the element have to be equal since it is assumed it moves only in the vertical plane but why does it have to be constant? Don't they just have to be equal?

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The fact that they have to be equal implies that the tension is constant (not necessarily in time, but along the length of the string), and here's why:

Let $T_x(t,x)$ denote the horizontal component of the tension as a function of time $t$ and position $x$ along the string. Let a small element of length $\Delta x$ of the string with left endpoint at position $x$ be given. As you point out, if the string element is to never accelerate horizontally, then the net tension in the $x$ direction must vanish at every instant in time $t$. By Newton's second law, this means that $$ T_x(t,x+\Delta x) - T_x(t,x) = 0 $$ dividing both sides by $\Delta x$ and then taking the limit $\Delta x \to 0$ shows that the derivative of the $x$-component of the tension with respect to $x$ must vanish; $$ \frac{\partial T_x}{\partial x}(x) = 0 $$ This implies that $T_x$ is constant in $x$. In particular, this means that there is some function $f$ for which $$ T_x(t, x) = f(t) $$ for all times $t$. The tension does not need to be constant in time, but it must be constant in position along the length of the string.

Physically, you could imagine that the tension in the string results from attaching weights to each end that are hanging, and if you want to change the horizontal tension in the string at a function of time, you just change the weights providing the tension as a function of time. If you were to change the tension as a function of time, however, then the wave speed along the string would depend on time, and this would make the wave equation harder to solve.

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That part makes sense, but could it not be a function of time? –  Shawn Apr 1 '13 at 17:46
    
@Shawn I made some edits to address this point. –  joshphysics Apr 1 '13 at 18:01
    
Thanks for the explanation –  Shawn Apr 1 '13 at 18:56
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