/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Tall buildings are designed to s... [FREE SOLUTION] | 91Ó°ÊÓ

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Tall buildings are designed to sway in the wind. In a \(100-\mathrm{km} / \mathrm{h}\) wind, for example, the top of the 110 -story Sears Tower oscillates horizontally with an amplitude of \(15 \mathrm{~cm}\). The building oscillates at its natural frequency, which has a period of \(7.0 \mathrm{~s}\). Assuming SHM, find the maximum horizontal velocity and acceleration expericnced by a Sears employee as she sits working at her desk located on the top floor. Compare the maximum acceleration (as a percentage) with the acceleration due to gravity.

Short Answer

Expert verified
The maximum velocity is approximately 0.1346 m/s, and the maximum acceleration is 0.1205 m/s², which is about 1.23% of gravity.

Step by step solution

01

Identify Known Variables

From the problem, we identify the following known variables:- Amplitude of oscillation, \( A = 15 \text{ cm} = 0.15 \text{ m} \)- Period of oscillation, \( T = 7.0 \text{ s} \)- Gravitational acceleration, \( g = 9.81 \text{ m/s}^2 \)
02

Calculate Angular Frequency

The angular frequency \( \omega \) can be calculated using the formula: \[ \omega = \frac{2\pi}{T} \]Substituting the period \( T = 7.0 \text{ s} \), we get:\[ \omega = \frac{2\pi}{7.0} \approx 0.897 \text{ rad/s} \]
03

Determine Maximum Velocity

The formula for maximum velocity \( v_{\text{max}} \) in simple harmonic motion is:\[ v_{\text{max}} = \omega A \]Substituting \( \omega \approx 0.897 \text{ rad/s} \) and \( A = 0.15 \text{ m} \):\[ v_{\text{max}} = 0.897 \times 0.15 \approx 0.1346 \text{ m/s} \]
04

Determine Maximum Acceleration

The formula for maximum acceleration \( a_{\text{max}} \) in simple harmonic motion is:\[ a_{\text{max}} = \omega^2 A \]Substituting \( \omega \approx 0.897 \text{ rad/s} \) and \( A = 0.15 \text{ m} \):\[ a_{\text{max}} = (0.897)^2 \times 0.15 \approx 0.1205 \text{ m/s}^2 \]
05

Percentage of Maximum Acceleration to Gravity

To find the percentage of maximum acceleration compared to gravity, use:\[ \text{Percentage} = \left(\frac{a_{\text{max}}}{g}\right) \times 100 \%\]Substituting \( a_{\text{max}} \approx 0.1205 \text{ m/s}^2 \) and \( g = 9.81 \text{ m/s}^2 \):\[ \text{Percentage} = \left(\frac{0.1205}{9.81}\right) \times 100 \approx 1.23\% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Frequency
Angular frequency, often symbolized as \( \omega \), is a fundamental concept in simple harmonic motion (SHM). This term measures how rapidly an object oscillates or cycles through its motion. It is typically expressed in radians per second (rad/s).
The formula to calculate angular frequency \( \omega \) is:
  • \( \omega = \frac{2\pi}{T} \)
where \( T \) is the period of oscillation, which is the time it takes for one complete cycle of motion.
In our example of the Sears Tower, the period \( T \) is given as 7 seconds. So, \( \omega = \frac{2\pi}{7} \approx 0.897 \text{ rad/s} \).
This means that the top of the building completes about 0.897 radians of its oscillation per second. Understanding angular frequency helps us describe the rhythm of motion and is essential for finding other important motion parameters like maximum velocity and acceleration.
Amplitude of Oscillation
The amplitude of oscillation is a term used to explain the maximum extent of displacement from the equilibrium position during oscillation. It is a crucial aspect of simple harmonic motion that defines the peak range or distance that the object travels from its center or resting state.
In simpler terms, amplitude indicates how far, at maximum, the object swings away from its median position during each cycle of motion. This is typically conveyed in meters or centimeters depending on the context.
In this scenario, the amplitude for the Sears Tower is 15 cm or 0.15 meters. This figure reflects how much the top of the tower sways to one side at its furthest reach. Why is understanding amplitude important? It helps us identify how dramatic or mild the sway is, which is pivotal in assessing the impact on the structure and the comfort of the people inside.
Maximum Acceleration
Maximum acceleration in simple harmonic motion refers to the highest rate of velocity change that occurs during oscillation. It gives us insight into the forces exerted on the object at the extremes of its motion.
For any object undergoing SHM, the formula to determine this maximum acceleration, \( a_{\text{max}} \), is:
  • \( a_{\text{max}} = \omega^2 A \)
where \( \omega \) is the angular frequency and \( A \) is the amplitude of oscillation.
In our building sway example, using \( \omega \approx 0.897 \text{ rad/s} \) and \( A = 0.15 \text{ m} \), we get \( a_{\text{max}} \approx 0.1205 \text{ m/s}^2 \).
Understanding maximum acceleration is crucial since it tells us how intense the forces can be during the oscillation. In structural engineering, it helps in comparing how those forces measure against gravitational acceleration (\( g = 9.81 \text{ m/s}^2 \)), allowing for a safety assessment as seen by the calculation of it as a percentage of gravity, here about 1.23%.

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Most popular questions from this chapter

(III) A mass \(m\) is at rest on the end of a spring of spring constant \(k .\) At \(t=0\) it is given an impulse \(J\) by a hammer. Write the formula for the subsequent motion in terms of \(m, k, J,\) and \(t\)

(II) A vertical spring with spring stiffness constant 305 \(\mathrm{N} / \mathrm{m}\) oscillates with an amplitude of 28.0 \(\mathrm{cm}\) when 0.260 \(\mathrm{kg}\) hangs from it. The mass passes through the equilibrium point \((y=0)\) with positive velocity at \(t=0 .\) (a) What equation describes this motion as a function of time? (b) At what times will the spring be longest and shortest?

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(II) A bungee jumper with mass 65.0 \(\mathrm{kg}\) jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 43.0 \(\mathrm{s}\) . He finally comes to rest 25.0 \(\mathrm{m}\) below the level of the bridge. Estimate the spring stiffness constant and the unstretched length of the bungee cord assuming SHM.

A \(0.650-\mathrm{kg}\) mass oscillates according to the equation \(x=0.25 \sin (5.50 t)\) where \(x\) is in meters and \(t\) is in seconds. Determine (a) the amplitude, (b) the frequency, (c) the period, (d) the total energy, and ( \(e\) ) the kinetic energy and potential energy when \(x\) is 15 \(\mathrm{cm} .\)

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