Physics Oscillation question-please help walk me understand what to do:)

Question

The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.15 Hz , and the acceleration of the top of the building can reach 1.8 % of the free-fall acceleration, enough to cause discomfort for occupants.

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

Answer 1

Let

f = 0.15 Hz be the frequency of swaying,

g = 9.81 m/s² be the free-fall acceleration,

a_max = 0.018×g be the maximum acceleration.

step 1)

In general, size of oscillation follows the formula

x(t) = Asin(2πft) (1)

where

x(t) is the displacement from the vertical at time t, and

A is the maximum such displacement.

(You are being asked to find 2×A)

step 2)

In general, the acceleration at time t follows the formula

a(t) = -(2πf)²Asin(2πft) (2)

If you had calculus you can derive that by differentiation equation (1) twice.

If you did not have calculus, you just need to accept that.

step 3)

Formula (2) implies that a(t) reaches its maximum when sin(2πft) = -1,

that is, the maximum acceleration is

a_max = (2πf)²A

step 4)

Using the given magnitude of the maximal acceleration

(2πf)²A = 0.018×g

Solving for A:

A = 0.018×g/(2πf)² = 0.018×9.81/(2×π×0.15)² ≈ 0.2 m

So the side-to-side swing is 0.4 m (independent of the height of the building).