# Magnitude of tension in a bent string

I understand tension in a straight string as a reaction force to a weight, which acts along the string, ultimately resulting from the attractive forces between the constituent particles of the string. I'm not sure how to understand tension in a curved string, such as a vibrating guitar string. I understand that an ideally elastic string is meant to have tension acting at tangent to the string, but that's about it. Is the magnitude of the tension vector equal throughout the string?

I've been going through various derivations of the wave equation for transverse motion of a string fixed at both ends and they all use the assumption that the tensions at each end of an infinitesimal segment of the string are equal in magnitude. I don't understand how this assumption is warranted.

The formula speed of wave propagation in a string involves this tension $T$ in which it is a constant. Can tension in a string be constant when it's oscillating?

Consider a small segment segment under tension

From the balance of the horizontal axis you have

$$-T \cos \theta + (T+{\rm d}T) \cos (\theta+{\rm d}\theta) = 0$$

$${\rm d}T \approx T \tan(\theta) {\rm d} \theta$$

Integrating by separation of variables

$$\int \frac{1}{T}\,{\rm d}T = \int \tan(\theta)\,{\rm d}\theta + K$$ $$T = \frac{{\rm e}^K}{\cos \theta}$$

So the tension varies by the slope. High slopes makes the tension higher.

• Thanks for the answer. Could I ask though what approximation are you using to get to the line $dT\approx Ttan(\theta)d\theta$? Nov 12, 2015 at 10:59
• But if indeed the tension at different points along a curved string changes, how can this be reconcilable with derivations of the wave equation for strings in which there is no longitudinal motion, so that the horizontal components of tension are equal, as in link? In this derivation they say that the horizontal component of tension must be constant throughout the string. But I don't see how the authors could make that inference without $T_1, T_2$ being constants themselves. Nov 12, 2015 at 11:18
• 1st order taylor series on ${\rm d}\theta$ can get you $$\cos \theta {\rm d}T - (T+{\rm d}T) \sin \theta {\rm d}\theta =0$$ and ignore any ${\rm d}\theta {\rm d}T$ terms. Nov 12, 2015 at 13:18
• Small angle approximation yields the tensions to be constants as ${\rm d}T \approx 0$, or since $\cos \theta \approx 1$ then $$T = {\rm e}^{K}$$ Nov 12, 2015 at 13:24
• Come to think of it, I think the above is wrong. I know the catenary solution of a hanging cable is $$T = H \cosh \left( \frac{x}{2 a} \right)$$ where $H$ is the minimum tension (at the low point of the cable), $a=\frac{H}{w}$ is the catenary constant and $w$ is the unit weight of the cable (in force/length units). This implies that $$T = H \sqrt{1+\theta^2}$$ Nov 12, 2015 at 13:45

No, certainly not. In linear model there is something like static tension $T_0$ and time dependent tension $T'$.

When there is for example a standing wave on a string with fixed ends, there is a minimum of string displacement (zero) but maximum of string tension $T'$ at the fixed ends. (As an effective value throught the time.)