Question

The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.18 Hz , and the acceleration of the top of the building can reach 1.9 % of the free-fall acceleration, enough to cause discomfort for occupants.

Required:
What is the total distance, side to side, that the top of the building moves during such an oscillation?

Answer 1

Answer:

The total distance, side to side, that the top of the building moves during such an oscillation is approximately 0.291 meters.

Explanation:

Let suppose that the building is experimenting a Simple Harmonic Motion due to the action of wind. First, we determine the angular frequency of the system ($\omega$), in radians per second:

$\omega = 2\pi\cdot f$ (1)

Where $f$ is the frequency, in hertz.

If we know that $f = 0.18\,hz$, then the angular frequency of the system is:

$\omega = 2\pi\cdot (0.18\,hz)$

$\omega \approx 1.131\,\frac{rad}{s}$

The maximum acceleration experimented by the system is represented by the following formula, of which we estimate amplitude of the oscillation:

$r\cdot g = \omega^{2}\cdot A$ (2)

Where:

$r$ - Ratio of real acceleration to free-fall acceleration, no unit.

$g$ - Free-fall acceleration, in meters per square second.

$A$ - Amplitude, in meters.

If we know that $\omega \approx 1.131\,\frac{rad}{s}$, $r = 0.019$ and $g = 9.807\,\frac{m}{s^{2}}$, then the amplitude of the oscillation is:

$A = \frac{r\cdot g}{\omega^{2}}$

$A = \frac{(0.019)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{\left(1.131\,\frac{rad}{s} \right)^{2}}$

$A \approx 0.146\,m$

The total distance, side to side, is twice the amplitude, that is to say, a value of approximately 0.291 meters.