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The above is a derivation for the wave speed equation in my physics textbook. However, I've read online that this equation is only true for waves with small amplitudes. I do not see where this assumption is made in the derivation, so why is the equation only true for small amplitudes?

enter image description here

The above picture shows the vertical restoring force should be 2*T*sin(phi)

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The explanation is not a very full one. As you correctly note, you're taking a limit, so the assumption $\sin\theta \to\theta$ as $\delta z\to0$ becomes exact. So Eq 16-23 contains no approximation.

The assumption creeps in subtly when one assumes that the force calculated in Eq 16-23 is at right angles to the $z$ axis. That is, that $\mathrm{d}y/\mathrm{d} z$ is small, so that the normal to the tangent to the curve stays approximately vertical in the diagram. The best way to understand all this is to work out a more accurate equation; then the vertical component of the force restoring the small length $\mathrm{d}\,s$ of string is

$$T\,\partial\theta \,\cos\theta = \mathrm{d}s\, T\,\frac{\partial\theta}{\partial s}\,\cos\theta = \mathrm{d}s\, T\,\frac{\partial^2y}{\partial z^2}\,\frac{1}{\left(1+\left(\frac{\partial y}{\partial z}\right)^2\right)^2}$$

(recalling that $\frac{\partial\theta}{\partial s}$ is the curvature of the string and then using the formula for the curvature) and THEN you approximate that $\frac{\partial y}{\partial z}\ll 1$ and, equivalently, that $\mathrm{d}z = \mathrm{d}s$. The small amplitude approximation is then indirect: we're directly assuming small gradients, which imply and are implied by small amplitudes, given that we know the wavelength is limited.

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  • $\begingroup$ How did you get the starting equation Tdthetacos(theta)? For the last step, should the denominator be raised to the power of 3/2? Did you replace cos(theta) with cos(0) since theta is small? Can't we make the force at right angles to the z-axis by making ds small? We know that the tangent at the middle is flat, so the tension force will be approximately flat if we stay near the middle, right? $\endgroup$ – roobee May 23 '16 at 18:42
  • $\begingroup$ The formula is simply the same as 16-23 in your text with a $\cos\theta$ added to account for the fact that the force is not perfectly vertical but skewed by the string's gradient. And you're right about the missing $3/2$ power, I've now put it back in all its glory (it multiplies the $\cos\theta$ factor to become the power of two now in the denominator on the RHS). Also, you can't make the force at right angles at all points: recall that the string is assumed to be perfectly bendable, which means that it can resist no shear and can only impart tension along its tangent. $\endgroup$ – WetSavannaAnimal May 24 '16 at 13:01
  • $\begingroup$ doesn't the sin(theta) in the textbook already account for the skew? $\endgroup$ – roobee Jun 3 '16 at 3:52
  • $\begingroup$ @roobee No, it's simply measuring the arclength. I should probably have used a different symbol from $\theta$; in the book its simply the angle subtended by the section of string at the center of curvature, and we should use a second symbol, say $\phi$, to measure the skew. $\endgroup$ – WetSavannaAnimal Jun 3 '16 at 4:15
  • $\begingroup$ I've added a picture in my question. I've replaced theta with phi. It shows that the vertical component of the force is 2*T*sin(phi), so doesn't that show that sin(phi) accounted for the skew? $\endgroup$ – roobee Sep 16 '16 at 17:49
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Right off the bat in eq (16-23) it's assumed that the restoring force is linear in the displacement. That's only true for small displacements.

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  • $\begingroup$ Are you referring to where sin(theta) changes to theta? If so, I thought that was because theta was small due to the segment being small, and not because the amplitude was small. $\endgroup$ – roobee May 23 '16 at 2:38
  • $\begingroup$ No, I'm not. I was referring to the approximation that $\tau$ is taken to be a constant, whereas we know that the tension will increase as the string is stretched. However, as you can now see, there's more to it than that. $\endgroup$ – garyp May 23 '16 at 11:49

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