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1 year ago

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

Physics

1 answer:

Jlenok [28]1 year ago

5 0

The total side to side distance at the top of the building, covered during such an oscillation is 8.4 cm.

Solution:

The height of building is, h = 152 m.

The frequency on windy days is, f = 0.17 Hz.

The acceleration on the top of building is, a = 2/100g (Here g is gravitational acceleration).

$a = A \times \omega ^{2}$

$a = A \times (2\pi f) ^{2}$

$\frac{2}{100} \times g = A \times (2\pi \times 0.17) ^ 2$

A = 0.042m

hence the total distance is twice the distance

that is 0.084m

<h3>What is oscillation?</h3>

The recurrent or periodic change of a quantity around a central value (often an equilibrium point) or between two or more different states is known as oscillation.
Alternating current and a swinging pendulum are two common examples of oscillation. Physics can employ oscillations to approximate complicated interactions, like those between atoms.

To learn more about Oscillation with the given link

brainly.com/question/17133973

#SPJ4

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?

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