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Chapter 12: Problem 51

The period of a simple harmonic oscillator is \(0.0125 \mathrm{~s}\). What is the frequency?

Short Answer

Expert verified
The frequency of the simple harmonic oscillator is \(80 \) Hz.

Step by step solution

01

Identify the Knowns

In the problem, we know that the period \( T \) of the simple harmonic oscillator is \( 0.0125 \) seconds.
02

Apply the Formula for Frequency

The formula for finding the frequency \( f \) when the period is known is \( f = \frac{1}{T} \). Substitute the given period \( T = 0.0125\) s into the formula.
03

Compute the Frequency

Calculate the frequency using the formula: \( f = \frac{1}{0.0125} = 80 \) Hz

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Key Concepts

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

Frequency
Frequency is a key concept in understanding simple harmonic motion (SHM). It represents the number of oscillations or cycles that occur in one second. In the context of our oscillator, we calculate frequency using the formula \[ f = \frac{1}{T} \]where \( f \) is the frequency and \( T \) is the period of the oscillation.
  • Unit of Frequency: The unit of frequency is hertz (Hz), which is equal to one cycle per second.
  • Relation to Period: Frequency is the inverse of the period, which makes it easy to calculate when the period is known.
In our example, the period is given as 0.0125 seconds. By substituting this value into the formula, we find that the frequency is 80 Hz. This means that the oscillator completes 80 cycles every second.
When learning about frequency in SHM, remember that a higher frequency indicates quicker oscillations, while a lower frequency suggests slower oscillations.
Period
The period of an oscillator is the time taken to complete one full cycle of its motion. In simple harmonic motion, the period is a crucial parameter as it defines how long the oscillator takes to return to its original position.
  • Symbol and Units: The period is denoted by \( T \) and is measured in seconds (s).
  • Significance: A shorter period means that the oscillator moves back and forth more quickly, while a longer period indicates a slower pace.
In the provided problem, the period is 0.0125 seconds. This short period tells us that the oscillator cycles very swiftly, completing a cycle in a very small amount of time.
Understanding the concept of the period is essential, as it helps explain how fast an oscillator, such as a pendulum or a spring-mass system, moves during its oscillations.
Oscillator
An oscillator in physics refers to a system that moves about an equilibrium position in a repeated, cyclical pattern. Simple harmonic oscillators are a fundamental type of oscillator, characterized by a restoring force that is proportional to their displacement.
  • Examples: Common examples include pendulums, vibrating strings on a guitar, and springs.
  • Nature of Motion: The motion of simple harmonic oscillators is sinusoidal in nature, meaning it can be described by sine or cosine functions.
  • Energy Conservation: Oscillators exhibit a constant exchange of potential and kinetic energy during their motion.
In our example problem, the oscillator's frequency and period were examined, showcasing its rapid, repetitive motion. Although simple in concept, oscillators form the basis for understanding more complex phenomena in physics.

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Most popular questions from this chapter

A simple harmonic oscillator completes 1250 cycles in \(20 \mathrm{~min}\). Calculate (a) the period and (b) the frequency of the motion.

The potential energy of an object on a spring is \(2.4 \mathrm{~J}\) at a location where the kinetic energy is \(1.6 \mathrm{~J}\). If the amplitude of the simple harmonic motion is \(20 \mathrm{~cm}\), (a) calculate the spring constant and (b) find the largest force that it experiences.

Estimate the period for normal, human eye blinking.

The acceleration of an object that has a mass of \(0.025 \mathrm{~kg}\) and exhibits simple harmonic motion is given by \(a(t)=\left(10 \mathrm{~m} / \mathrm{s}^{2}\right) \cos (\pi t+\pi / 2)\). Calculate its velocity at \(t=2 \mathrm{~s}\), assuming the object starts from rest at \(t=0\). SSM

A \(110-\mathrm{g}\) pendulum bob swings at the end of a \(15.0\)-m-long wire. The pendulum's damping coefficient is \(0.010 \mathrm{~kg} / \mathrm{s}\). What is the amplitude of the pendulum after 3 oscillations if the initial amplitude is \(1.5 \mathrm{~m}\) ? SSM
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