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Chapter 15: Problem 10

An oscillating block-spring system takes \(0.75 \mathrm{~s}\) to begin repeating its motion. Find (a) the period, (b) the frequency in hertz, and (c) the angular frequency in radians per second.

Short Answer

Expert verified
(a) Period: 0.75 s, (b) Frequency: 1.33 Hz, (c) Angular frequency: 8.38 rad/s.

Step by step solution

01

Understand the Problem Statement

The problem describes a block-spring system in oscillation. The time taken for the system to begin repeating its motion is given as 0.75 seconds. This time is known as the period of the motion.
02

Determine the Period

The period is the duration of time it takes for one complete cycle of the oscillation. Given that the system takes 0.75 seconds to begin repeating its motion, the period \( T \) of the system is \( T = 0.75 \) seconds.
03

Calculate the Frequency

The frequency \( f \) is the number of complete cycles per second. It is the reciprocal of the period. Therefore, the frequency is given by the formula \( f = \frac{1}{T} \). Substituting the known period: \( f = \frac{1}{0.75} = \frac{4}{3} \approx 1.33 \) Hz.
04

Determine the Angular Frequency

The angular frequency \( \omega \) is the rate of change of the angle with respect to time in a circular motion, which can also be applied to oscillations. It is related to the frequency by the formula \( \omega = 2\pi f \). Substituting the frequency from the previous step: \( \omega = 2\pi \times \frac{4}{3} \approx 8.38 \) radians per second.

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Key Concepts

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

Period
In the world of oscillations, the period is a central concept. It refers to the time it takes for an oscillating system to complete one full cycle of motion. When you think about this concept, imagine a pendulum swinging back and forth, or a block springing up and down.
  • The period is simply the time from one peak to the next, or from one trough to the next.
  • We measure period in seconds, and it's denoted as \( T \).
For example, if a block-spring system takes 0.75 seconds to complete a cycle and return to its starting position, we determine that the period \( T = 0.75 \) seconds. Understanding the period is the first step to delving deeper into oscillatory motion.
Frequency
Another key player in understanding oscillations is frequency. While the period tells us how long one cycle takes, frequency tells us how many cycles occur in one second. Simply put, frequency is the inverse of the period.

  • If the period is long, the frequency will be lower because fewer cycles happen each second.
  • Conversely, if the period is short, the frequency is higher, meaning more cycles happen per second.
We express frequency in hertz (Hz), which is equivalent to cycles per second.
Given our block-spring system example with a period of 0.75 seconds, the frequency \( f \) can be calculated using the formula \( f = \frac{1}{T} \). Plugging the value in, we find \( f = \frac{1}{0.75} \approx 1.33 \) Hz.
Understanding frequency is crucial, as it gives insight into the rate at which oscillations occur.
Angular Frequency
Angular frequency takes the idea of frequency one step further. It is particularly useful in contexts involving circular motion and oscillations, where angles are relevant. Angular frequency, denoted by \( \omega \), describes how quickly the angle changes in radians per second.
  • It is related to the regular frequency \( f \) by the formula \( \omega = 2\pi f \).
  • This relationship stems from the fact that one complete cycle of circular motion corresponds to an angle of \( 2\pi \) radians.
In our block-spring system, with a frequency of approximately 1.33 Hz, the angular frequency can be calculated as \( \omega = 2\pi \times 1.33 \approx 8.38 \) radians per second.
Understanding angular frequency is essential for studying systems that have rotational aspects or involve sinusoidal functions. It's a crucial concept for analyzing oscillatory systems comprehensively.

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Most popular questions from this chapter

A uniform circular disk whose radius \(R\) is \(12.6 \mathrm{~cm}\) is suspended as a physical pendulum from a point on its rim. (a) What is its period? (b) At what radial distance \(r

At a certain harbor, the tides cause the ocean surface to rise and fall a distance \(d\) (from highest level to lowest level) in simple harmonic motion, with a period of \(12.5 \mathrm{~h}\). How long does it take for the water to fall a distance \(0.250 d\) from its highest level?

A \(3.0 \mathrm{~kg}\) particle is in simple harmonic motion in one dimension and moves according to the equation $$ x=(5.0 \mathrm{~m}) \cos [(\pi / 3 \mathrm{rad} / \mathrm{s}) t-\pi / 4 \mathrm{rad}] $$ with \(t\) in seconds. (a) At what value of \(x\) is the potential energy of the particle equal to half the total energy? (b) How long does the particle take to move to this position \(x\) from the equilibrium position?

A \(2.0 \mathrm{~kg}\) block is attached to the end of a spring with a spring constant of \(350 \mathrm{~N} / \mathrm{m}\) and forced to oscillate by an applied force \(F=\) \((15 \mathrm{~N}) \sin \left(\omega_{d} t\right),\) where \(\omega_{d}=35 \mathrm{rad} / \mathrm{s} .\) The damping constant is \(b=\) \(15 \mathrm{~kg} / \mathrm{s} .\) At \(t=0,\) the block is at rest with the spring at its rest length. (a) Use numerical integration to plot the displacement of the block for the first \(1.0 \mathrm{~s}\). Use the motion near the end of the \(1.0 \mathrm{~s}\) interval to estimate the amplitude, period, and angular frequency. Repeat the calculation for (b) \(\omega_{d}=\sqrt{k / m}\) and \((\mathrm{c}) \omega_{d}=20 \mathrm{rad} / \mathrm{s}\)

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance \(d\) from the \(50 \mathrm{~cm}\) mark. The period of oscillation is \(2.5 \mathrm{~s} .\) Find \(d\)
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