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I’m confused about the force that is experienced by a small piece of a continuous massive spring from the rest of the spring to its left and right. On the one hand, the wave equation says the force is (in general) nonzero. On the other hand, it seems that the forces from either side of the small piece are both proportional to the space derivative of the wave, with opposite signs, so they should cancel to zero.

In more detail:

On the one hand, my understanding of one possible derivation of the wave equation is as follows. Start with a series of blocks each with mass $m$ laid out from left to right. Every two consecutive blocks are connected through a massless spring with spring constant $k_a$, and separated by a distance of $a$. $\psi(x,t)$ denotes the displacement at time $t$ of the block whose equilibrium position is $x$. Consider the block with equilibrium position $z$. The force it experiences from its left is $$k_a[\psi(z-a)-\psi(z)]\tag{1}$$ and the force it experiences from its right is $$k_a[\psi(z+a)-\psi(z)]\tag{2}$$ So its acceleration is $$\frac{\partial^2\psi}{\partial t^2}(z)=\frac{k_a}{m}[\psi(z-a)-2\psi(z)+\psi(z+a)]\tag{3}$$ Taylor expansion of $\psi(z-a)$ and $\psi(z+a)$ around $z$ gives $$\frac{\partial^2\psi}{\partial t^2}(z)=\frac{k_a}{m}[a^2\frac{\partial^2\psi}{\partial x^2}(z)+O(a^3)]\tag{4}$$ Taking the limit $a\to0$ while keeping $k_a a$ and $\frac{a}{m}$ constant, this becomes $$\frac{\partial^2\psi}{\partial t^2}(z)=k_aa\frac{a}{m}\frac{\partial^2\psi}{\partial x^2}(z)\tag{5}$$ This is the wave equation for the continuous massive spring. Here, the piece of the spring at $z$ is experiencing (in general) nonzero force, proportional to $\frac{\partial^2\psi}{\partial x^2}(z)$.

On the other hand, when deriving boundary conditions for the massive spring, one sometimes does something like the following. First, consider that the force experienced by the piece of spring at $z$ from its right is $k_aa\frac{\partial\psi}{\partial x}(z)$ (which comes from (2) after taking the limit $a\to0$). Then match this to the force specified by the system. For example, if the right end of a massive spring is at $y$, and it attaches to a wall through a Hooke’s Law spring with spring constant $k_c$, then the boundary condition would be $k_aa\frac{\partial\psi}{\partial x}(y)=-k_c\psi(y)$.

What I don’t understand is: consider the piece of spring at $z$, if the force from its right is $k_aa\frac{\partial\psi}{\partial x}(z)$, and the force from its left is $-k_aa\frac{\partial\psi}{\partial x}(z)$ (from (1) after taking the limit $a\to0$), then the net force would always be 0, which contradicts what the wave equation says. And what really confuses me is that both conclusions come from adding up (1) and (2) and taking the limit $a\to0$.

Wujie
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1 Answers1

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I think that is better instead of using massless springs, to write an equilibrium equation of an elastic body under uniaxial tension:

$$\Delta F = A(\sigma_{x+a}-\sigma_{x}) = m\frac{\partial^2 \psi}{\partial t^2}$$

Dividing by the volume:

$$\frac{\sigma_{x+a}-\sigma_{x}}{\Delta x} = \rho\frac{\partial^2 \psi}{\partial t^2}$$

As $$\sigma_x = E\epsilon_x = E\frac{\partial \psi}{\partial x}$$

where $E$ is the elasticity modulus. When $\Delta x -> 0$:

$$\frac{\partial^2 \psi}{\partial x^2} = \frac{\rho}{E}\frac{\partial^2 \psi}{\partial t^2}$$

It is a second order differential equation, and it is necessary 2 boundary conditions. One of them can a constant force, for example at right: $F_{x_R} = AE\frac{\partial \psi}{\partial x}|_{x_R}$

The second can be an opposite force at left: $F_{x_L} = -AE\frac{\partial \psi}{\partial x}|_{x_L}$

It doesn't mean that $F$ is constant through the spring. For any small element the assumption made in the deduction remains valid, and only at the ends of the spring there are constant forces.

Claudio Saspinski
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