Problem 5
Question
\(\bullet\) The frequency of a simple harmonic oscillator is doubled from \(0.25 \mathrm{~Hz}\) to \(0.50 \mathrm{~Hz}\). What is the change in its period?
Step-by-Step Solution
Verified Answer
The period decreases by 2 seconds.
1Step 1: Identify the original frequency and period
The original frequency of the simple harmonic oscillator is given as \( f_1 = 0.25 \, \mathrm{Hz} \). The period \( T \) is the reciprocal of the frequency, so the initial period \( T_1 \) is calculated as follows:\[ T_1 = \frac{1}{f_1} = \frac{1}{0.25} = 4 \, \mathrm{seconds} \]
2Step 2: Calculate the new period with the doubled frequency
The frequency is doubled from \( f_1 = 0.25 \, \mathrm{Hz} \) to \( f_2 = 0.50 \, \mathrm{Hz} \). The new period \( T_2 \) is the reciprocal of the new frequency:\[ T_2 = \frac{1}{f_2} = \frac{1}{0.50} = 2 \, \mathrm{seconds} \]
3Step 3: Determine the change in period
To find the change in the period, calculate the difference between the new period \( T_2 \) and the original period \( T_1 \):\[ \Delta T = T_2 - T_1 = 2 \, \mathrm{seconds} - 4 \, \mathrm{seconds} = -2 \, \mathrm{seconds} \]
4Step 4: Conclusion: Interpret the result
The change in the period is \(-2 \, \mathrm{seconds}\), meaning that the period of oscillation decreased by 2 seconds when the frequency was doubled.
Key Concepts
FrequencyPeriodOscillation
Frequency
In the context of a simple harmonic oscillator, the term **frequency** refers to how often the oscillation completes one full cycle in a second. It is measured in hertz (Hz), where 1 Hz is equal to one complete cycle per second.
Understanding frequency is key to comprehending vibrational motion or oscillations as it represents the rate of repetition or the speed of cycles occurring in a system. For example, if a system oscillates to and fro 5 times in one second, its frequency is 5 Hz.
Frequency can change, and understanding its relationship to other aspects of a system, such as period, is crucial. When the frequency increases, it means oscillations happen more often over the same time span. This change can affect other parameters involved in the oscillator's dynamics, which we'll explore below.
Understanding frequency is key to comprehending vibrational motion or oscillations as it represents the rate of repetition or the speed of cycles occurring in a system. For example, if a system oscillates to and fro 5 times in one second, its frequency is 5 Hz.
Frequency can change, and understanding its relationship to other aspects of a system, such as period, is crucial. When the frequency increases, it means oscillations happen more often over the same time span. This change can affect other parameters involved in the oscillator's dynamics, which we'll explore below.
Period
The **period** of a simple harmonic oscillator is the duration it takes to complete one full cycle of oscillation. Mathematically, it is the inverse of frequency and is denoted by the symbol **T**.
The formula is straightforward: \[ T = \frac{1}{f} \] Where \(T\) is the period, and \(f\) is the frequency. From this equation, it's clear that the period and frequency are inversely related.
As shown in our original exercise, if the frequency of an oscillator doubles, the period halves. For example, if an oscillator originally had a frequency of 0.25 Hz, its period was calculated to be 4 seconds. By doubling the frequency to 0.50 Hz, the period decreased to 2 seconds.
Understanding this relationship helps predict how oscillation behavior changes in response to adjustments in frequency, highlighting the flexibility and dynamic nature of oscillating systems.
The formula is straightforward: \[ T = \frac{1}{f} \] Where \(T\) is the period, and \(f\) is the frequency. From this equation, it's clear that the period and frequency are inversely related.
As shown in our original exercise, if the frequency of an oscillator doubles, the period halves. For example, if an oscillator originally had a frequency of 0.25 Hz, its period was calculated to be 4 seconds. By doubling the frequency to 0.50 Hz, the period decreased to 2 seconds.
Understanding this relationship helps predict how oscillation behavior changes in response to adjustments in frequency, highlighting the flexibility and dynamic nature of oscillating systems.
Oscillation
An **oscillation** is a repetitive back-and-forth motion around an equilibrium point, characteristic of systems undergoing simple harmonic motion.
Oscillations are observable in various contexts such as the swing of a pendulum, vibrations of a guitar string, or even the ebb and flow of ocean tides. What these have in common is the rhythmic motion that each unit undergoes over a periodic interval.
Oscillations are described not just by frequency and period, but also by amplitude, which is the maximum displacement from the equilibrium. However, in the exercise provided, frequency and period were central to understanding the change in oscillation dynamics when adjustments are made.
When the frequency increases in any oscillating system, the oscillation must complete more cycles in the same period, thereby altering how quickly the system returns to its equilibrium point. Conversely, the period shows how long each oscillation takes, providing a period-to-frequency feedback that describes motion continuity and consistency.
Oscillations are observable in various contexts such as the swing of a pendulum, vibrations of a guitar string, or even the ebb and flow of ocean tides. What these have in common is the rhythmic motion that each unit undergoes over a periodic interval.
Oscillations are described not just by frequency and period, but also by amplitude, which is the maximum displacement from the equilibrium. However, in the exercise provided, frequency and period were central to understanding the change in oscillation dynamics when adjustments are made.
When the frequency increases in any oscillating system, the oscillation must complete more cycles in the same period, thereby altering how quickly the system returns to its equilibrium point. Conversely, the period shows how long each oscillation takes, providing a period-to-frequency feedback that describes motion continuity and consistency.
Other exercises in this chapter
Problem 3
\(\bullet\) A \(0.75-\mathrm{kg}\) object oscillating on a spring completes a cycle every \(0.50 \mathrm{~s}\). What is the frequency of this oscillation?
View solution Problem 4
\(\bullet\) A particle in simple harmonic motion has a frequency of \(40 \mathrm{~Hz}\). What is the period of this oscillation?
View solution Problem 6
\(\bullet\) An object of mass \(0.50 \mathrm{~kg}\) is attached to a spring with spring constant \(10 \mathrm{~N} / \mathrm{m}\). If the object is pulled down \
View solution Problem 7
\(\bullet\) An object of mass \(1.0 \mathrm{~kg}\) is attached to a spring with spring constant \(15 \mathrm{~N} / \mathrm{m}\). If the object has a maximum spe
View solution