/*! 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 14 The top of a skyscraper sways ba... [FREE SOLUTION] | 91Ó°ÊÓ

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The top of a skyscraper sways back and forth, completing 95 full oscillation cycles in 10 minutes. Find (a) the period and (b) the frequency (in Hz) of its oscillatory motion.

Short Answer

Expert verified
The period of the skyscraper's oscillatory motion is 6.32 seconds and the frequency is 0.16 Hz.

Step by step solution

01

Compute the total time in seconds

The exercise has given the time period in minutes, thus, it will be converted into seconds because the standard unit of time in physics is seconds. So, 10 minutes is equal to \(10 \times 60 = 600\) seconds.
02

Calculate the Period

The period (T) is the total time divided by the number of cycles. Therefore the period is \(T = \frac{time}{cycles} = \frac{600 s}{95} = 6.32 s \)
03

Calculate the Frequency

Frequency (f) is the reciprocal of the period. It measures the number of cycles per second. So, the frequency is \(f = \frac{1}{T} = \frac{1}{6.32 s} = 0.16 Hz \)

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Key Concepts

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

Period
When we talk about the period of oscillatory motion, we are referring to the time it takes for one full cycle of motion to occur. In this context, a cycle is one full swing back and forth. Calculating the period is crucial in understanding how an oscillating object behaves over time.

To find the period (T), you need two key pieces of information: the total time taken for the oscillations and the number of oscillation cycles completed. The period can be calculated using the formula: \[ T = \frac{\text{total time}}{\text{number of cycles}} \]For the skyscraper in the given problem, it sways back and forth 95 times in 600 seconds. By dividing 600 seconds by 95 cycles, we find that each cycle takes about 6.32 seconds. This means that the period of the skyscraper's oscillation is 6.32 seconds.

Understanding the period helps in predicting when the swaying will repeat itself, which can be especially important in engineering and architecture to ensure the building's design can safely handle these movements.
Frequency
Frequency is an essential concept in oscillatory motion, as it tells us how often cycles occur in a given timeframe. Frequency is measured in hertz (Hz), which is equal to one cycle per second.

To find the frequency (f) from the period (T), we take the reciprocal of the period: \[ f = \frac{1}{T} \]In our example, with a period of 6.32 seconds, the frequency of the skyscraper's oscillation is approximately 0.16 Hz. This indicates that about 0.16 cycles are completed each second.

Knowing the frequency is helpful because it provides insight into the dynamics of motion. This can be particularly useful in various applications, such as tuning musical instruments or designing systems that need to respond to specific frequencies.
Physics Problem Solving
Solving physics problems involves a systematic approach that helps break down complex scenarios into manageable steps. For oscillatory motion problems like the one involving the swaying skyscraper, following a clear methodology is key.

Here’s a simple problem-solving strategy:
  • Understand the problem: Read the problem carefully to identify what is being asked and what information is given.
  • List the known values: In the skyscraper example, these include the time frame (10 minutes) and the number of oscillations (95 cycles).
  • Convert units if necessary: Always work within standard units for consistency, such as seconds for time in physics.
  • Apply relevant formulas: Use formulas like \( T = \frac{\text{total time}}{\text{number of cycles}} \) for period and \( f = \frac{1}{T} \) for frequency.
  • Check your calculations: Verify each step and ensure the answer makes logical sense in the context of the problem.
Following these steps not only aids in finding the correct answer but also enhances your understanding of the underlying concepts. This systematic approach can be applied across various types of physics problems, making it a versatile tool in your problem-solving toolbox.

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Most popular questions from this chapter

What will happen to the period of a mass-spring system if it's placed in a jetliner accelerating down a runway? What will happen to the period of a pendulum in the same situation?

A simple pendulum of mass \(m\) is swinging with period \(T\) and amplitude \(\theta_{\max }\). Find expressions for (a) its total energy and (b) its maximum speed.

This problem explores the nonlinear pendulum discussed P qualitatively in Conceptual Example 13.1. You can tackle this problem if you have experience with your calculator's differential-equation solving capabilities or if you've used a software program like Mathematica or Maple that can solve differential equations numerically. On page 252 we wrote Newton's law for a pendulum as \(1 d^{2} \theta / d t^{2}=-m g L \sin \theta\). (a) Rewrite this equation in a form suitable for a simple pendulum, but without making the approximation \(\sin \theta \cong \theta\). Although it won't affect the form of the equation, assume that your pendulum uses a massless rigid rod rather than a string, so it can turn completely upside down without collapsing. (b) Enter your equation into your calculator or software, and produce graphical solutions to the equation for the situation where you specify the initial kinetic energy \(K_{0}\) when the pendulum is at its bottommost position. In particular, describe solutions for (i) \(K_{0}\) \(K_{0} \leqslant U_{\max }\), and (iii) \(K_{0}>U_{\max } .\) Here \(U_{\max }\) is the maximum pos- (ii) \(K_{0} \leqslant U_{\max }\), and (iii) \(K_{0}>U_{\max } .\) Here \(U_{\max }\) is the maximum pos- sible potential energy for the system, which occurs when the pendulum is completely upside down; \(U_{0}=2 L m g\), where \(L\) is the pendulum's length.

A physics student, bored by a lecture on simple harmonic moion, idly picks up his pencil (mass \(8.65 \mathrm{~g}\), length \(18.8 \mathrm{~cm}\) ) by the tip with his frictionless fingers, and allows it to swing back and forth with small amplitude. If the pencil completes 5974 full cycles during the lecture, how long does the lecture last?

A wheel rotates at \(720 \mathrm{rpm}\). Viewed from the edge, a point on the wheel appears to undergo simple harmonic motion. What are (a) the frequency in \(\mathrm{Hz}\) and (b) the angular frequency for this SHM?

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