Why do tall buildings have low resonant frequencies?

Question

I know that tall buildings have low natural frequencies, hence they're more vulnerable to earthquakes, but why do they have low natural frequencies?

Answer 1

We can model a short building as a uniform cuboid of density $\rho$ occupying the region

$$0 \le x \le L_x$$

$$0 \le y \le L_y$$

$$0 \le z \le L_z$$

with its mass given by

$$M = \rho V = \rho A L_z = \rho L_x L_y L_z$$

The building is attached firmly to the ground ($xy$-plane).

Ignoring gravity and compressive stress, consider only the effects of the shear stress acting on the building.

Let the displacement of the building be in the positive $x$ direction by a uniform shear stress.

[![Image of building under shear stress][1]][1] [1]: https://i.sstatic.net/7MmVe.png

The shear strain, $\theta$, is related to the shear stress, $\frac{ F_{//} }{A}$, according to the equation

$$\frac{ F_{//} / A }{ \theta } = S$$

where $S$ is the shear modulus and is a constant for small $\theta$.

A restoring force that is directly proportional to its displacement suggests simple harmonic motion (SHM).

The position of a point on the building (by geometry) is

$$x = x_0 + z_0 \tan \theta$$ $$y = y_0$$ $$z = z_0$$

Since the displacement is assumed to be small, we can use the first order approximation for $\tan \theta$.

$$x \approx x_0 + z_0 \theta$$

To obtain the corresponding velocity, take the derivative of the position with respect to time.

$$v_x = z_0 \frac{d\theta}{dt}$$ $$v_y = 0$$ $$v_z = 0$$

Next, we try to compute the kinetic energy of the building. For a small mass element with neutral position at $(x_0,y_0,z_0)$ occupying a volume of $dV$, its kinetic energy is given by

\begin{align} dK &= \frac{1}{2} v^2 dM \\ &= \frac{1}{2} v_x^2 \rho dV \\ &= \frac{1}{2} (z_0 \frac{d\theta}{dt})^2 \rho dV \end{align}

For a sheet of mass element at height $z$, and occupying a thickness of $dz$, its kinetic energy is

$$dK = \frac{1}{2} z^2 (\frac{d\theta}{dt})^2 \rho (Adz) $$

The total kinetic energy of the building is given by

\begin{align} K &= \frac{1}{2} (\frac{d\theta}{dt})^2 \rho A \int_0^{L_z} z^2 dz \\ &= \frac{1}{6} (\frac{d\theta}{dt})^2 \rho A L_z^3 \end{align}

Next we calculate the potential energy described by the conservative shear force

$$\mathbf{F_{//}}(\theta) = - S A \theta \hat{\mathbf{x}}$$

\begin{align} U(\theta) &= - \int \mathbf{F_{//}} \cdot d \mathbf{x} \\ &= - \int_0^\theta (-S A \phi) L_z d\phi \\ &= \frac{1}{2} S A L_z \theta^2 \end{align}

Since energy is conserved, let the total energy be $E$

$$K + U = E$$ $$\frac{1}{6} (\frac{d\theta}{dt})^2 \rho A L_z^3 + \frac{1}{2} SAL_z \theta^2 = E$$ $$(\frac{d\theta}{dt})^2 + \frac{3S}{L_z^2 \rho} \theta^2 = \frac{6E}{M L_z^2}$$

This is an SHM differential equation and it’s natural (angular) frequency is given by

\begin{align} \omega_0 &= \sqrt{\frac{3S}{L_z^2\rho}} \\ &= \frac 1 L_z \sqrt{\frac{3S}{\rho}} \\ &= 2 \pi f_0 \end{align}

$f_0$ and $L_z$ are inversely proportional to each other. Therefore, a high building will have low resonant frequency.

Answer 2

In order to excite a resonance in something, you need to produce oscillations that add coherently, or in phase. This means that when you vibrate on object, you will want the reflections of the vibration to add with the new vibrations that are coming in. These reflections will take time to get from one end of the object to the other given the finite speed of sound. The taller the building is, the longer it takes for vibrations to get from the bottom to the top of the building. Therefore, a lower frequency wave is required to excite the resonance.

Answer 3

Generally speaking mechanical structures, if they 'ring' at all, will ring at frequencies determined by the properties of stiffness (elasticity) and mass. The frequency in most cases increases with increasing stiffness, but decreases with increase in mass.

Buildings have considerable stiffness, but not necessarily that large considering the relative mass involved. So it's fundamentally the relative stiffness to mass that determines the low frequencies one observes in buildings.

Answer 4

Excellent answers have been given above, so there may be considerable overlap with mine.

Consider a mass $m$ at the end of a spring with spring constant $k$. Admittedly, this is a very crude model but physicists have a way of making simple models. Let's call it a lumped 1D model of a building. The resonance frequency is $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$. For a taller building the spring constant $k$ goes down like $1/h$. At the same time the mass $m$ increases with $h$. So in this very simple model the frequency scales with $1/h$.

Using a tuned mass damper is one way to deal with this.

Answer 5

Real pendulums are pendulums with non-point distribution of mass. For these, to calculate time period as a function of length, we can apply the regular formula for harmonic motion after considering the length of the oscillator to be the the distance between the pivot and the center of mass. A tall building would obviously have pretty high center of mass, hence it'll have a very low frequency. It's like considering a really long simple pendulum, though the equations are somewhat different.

---

More specifically, we can model the building as a standing wave with one open end. The equation for position in a standing wave is $$y=2A\cos(\omega t)\sin(\frac{2\pi x}{L})$$where $L$ is the total length of the string, $x$ is the distance of the particular point from the end of the spring, and $$\omega = 2\pi f$$ Also, $${f=\frac{1}{2L}\times k'}$$ $k'$ is dependent upon mass per unit length of the body and tension. $L$ is the length of the whole body.

Clearly, there's an inverse proportion between frequency and length, so a tall building will have a low frequency.