# Transversal wave speed derivation for small amplitudes

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?

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

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.

• 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? – roobee May 23 '16 at 18:42
• 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. – WetSavannaAnimal May 24 '16 at 13:01
• doesn't the sin(theta) in the textbook already account for the skew? – roobee Jun 3 '16 at 3:52
• @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. – WetSavannaAnimal Jun 3 '16 at 4:15
• 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? – roobee Sep 16 '16 at 17:49

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.

• 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. – roobee May 23 '16 at 2:38
• 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. – garyp May 23 '16 at 11:49