Question

The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.17 Hz, and the acceleration of the top of the building can reach 2.0% of the free-fall acceleration, enough to cause discomfort for occupants. What is the total distance, side to side, that the top of the building moves during such an oscillation?

Answer 1

Answer:

d = 8.4 cm

Explanation:

In order to calculate the amplitude of oscillation of the top of the building, you use the following formula for the max acceleration of as simple harmonic motion:

$a_{max}=A\omega^2$           (1)

A: amplitude of the oscillation

w: angular speed of the oscillation = 2$\pi$f

f: frequency = 0.17Hz

The maximum acceleration of the top of the building is a 2.0% of the free-fall acceleration. Then, you have:

$a_{max}=0.02(9.8m/s^2)=0.196\frac{m}{s^2}$

Then, you solve for A in the equation (1) and replace the values of the parameters:

$A=\frac{a_{max}}{\omega^2}=\frac{a_{max}}{4\p^2i f^2}\\\\A=\frac{0.196m/s^2}{16\pi^2(0.17Hz)^2}\\\\A=0.042m=4.2cm$

The total distance, side to side, of the oscilation of the top of the building is twice the amplitude A. Then you obtain:

d = 2A = 2(4.2cm) = 8.4cm