# Frequency of longitudinal waves on a string

I recently heard the composer Ellen Fullman on her Long String Instrument, which consists of steel wires of approx. 20 m length played by rubbing/stroking the string longitudinally. Changing the tension of the string while playing does not affect its frequency but just increases the intensity, unlike normal transverse vibration which goes up with increased tension. Does anyone know why this is the case? What is the relationship of frequency to the other parameters?

If a string of length $$\ell$$ is clamped firmly at both ends, it can support standing (stationary) waves of fundamental frequency $$f_0=v/2 \ell,$$ in which $$v$$ is the speed of wave propagation along the string. So it is the how the speeds of the two types of wave depend on tension that determines how the frequencies depend on tension.

The speed of a wave depends on the accelerations of portions of the string, when disturbed.

For transverse waves, the transverse acceleration of a small portion of string depends on the difference in transverse components of the string's tension either side of the portion. The transverse components arise because of differences in transverse displacements of the string making the string at different angles to its undisplaced direction. These transverse force components are proportional to the tension, $$T,$$ itself (which we can show to be hardly changed anywhere along the string by the passage of the wave, provided its amplitude is small compared to the wavelength). By following through this argument we arrive at a famous equation for the speed of transverse waves on a string of mass $$\mu$$ per unit length:$$v_\text{tr}=\sqrt{\frac{T}{\mu}}$$

For longitudinal waves, the acceleration of a small portion of the string depends on the difference in local tension in the string either side of that portion. This difference doesn't depend on $$T,$$ the mean tension or tension when the spring is undisturbed. Instead it depends on the string modulus, $$\lambda$$ (change in tension per unit fractional stretching), as well as the difference in displacements of either end of a small portion of string. The speed of longitudinal waves is given approximately by $$v_\text{lo}=\sqrt{\frac{\lambda}{\mu}}.$$

• Note that $\mu$ itself would decrease ever so slightly as you increase the tension, but it's unlikely that this would cause a change perceptible to the ear. – Michael Seifert Jun 19 at 14:33
• Wow fantastic and clear response! Thank you very much! So frequency still depends on the density of the string but not on the average tension arrived through stretching from end to end. As I understand, the modulus $\lambda$ is a constant for a given material? In this case, the fundamental frequency of a string with length $L$ vibrating longitudinally is: $$f_0=\frac{1}{2L}\sqrt{\frac{\lambda}{\mu}}$$ – Thomas Nicholson Jun 20 at 12:43
• Agree. Thank you for the question – it made me think! I called $\lambda,$ the increase in tension per unit fractional extension, the 'modulus' after the spring modulus of the applied mathematics of my schooldays. But as well as depending on the material of the string, $\lambda,$ will be proportional to the string cross-sectional area, $A.$. We could write $\lambda=\lambda'A,$ in which $\lambda'$ depends on just the material. $\mu,$ the mass per unit length, is of course given by $\mu=\rho A,$ in which $\rho$ is the density. So we have the alternative from, $v=\sqrt (\lambda'/\rho).$ – Philip Wood Jun 20 at 14:37
• Actually this is very insightful, thank you. Just doing dimensional analysis was useful for me: I guess the modulus $\lambda$ would be measured in Pascals ... it would have to be given for the entire cross-sectional area $A$ in $m^2$ to work out as a velocity in the end: $$\sqrt{\frac{kg}{m \cdot s^2} \cdot m^2 \cdot \frac{m}{kg}} = \frac{m}{s}$$ Does this seem to make sense to you? – Thomas Nicholson Jun 20 at 18:30
• Took me a minute or two to figure out what was going on under the square root, but it's fine. In $v=\sqrt(\lambda/\mu,$ the modulus, $\lambda,$ has the units of force, because it's the force per unit $fractional$ extension. But you knew that! – Philip Wood Jun 20 at 19:28