A particle is executing simple harmonic motion. Which one of the following statements about the acceleration of the oscillating particle is true?
It is proportional to the frequency of oscillation
Understanding Acceleration in Simple Harmonic Motion (SHM)
Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to the displacement. This leads to an acceleration that is also proportional to the displacement and directed towards the equilibrium position.
Key Equations for SHM
For a particle executing SHM along the x-axis:
- Displacement: \(\text{x}(\text{t}) = \text{A} \cos(\omega \text{t} + \phi)\)
- Velocity: \(\text{v}(\text{t}) = \frac{\text{dx}}{\text{dt}} = -\text{A}\omega \sin(\omega \text{t} + \phi)\)
- Acceleration: \(\text{a}(\text{t}) = \frac{\text{dv}}{\text{dt}} = -\text{A}\omega^2 \cos(\omega \text{t} + \phi)\)
Here, \(\text{A}\) is the amplitude, \(\omega\) is the angular frequency, \(\text{t}\) is time, and \(\phi\) is the initial phase angle.
From the equations, we can see that \(\text{a}(\text{t}) = -\omega^2 \text{x}(\text{t})\). This is the defining characteristic of SHM: acceleration is proportional to displacement (\(\text{x}\)) and is directed towards the equilibrium position (due to the negative sign and \(\text{a}\) being opposite in direction to \(\text{x}\)).
The angular frequency \(\omega\) is related to the linear frequency \(\text{f}\) by \(\omega = 2\pi\text{f}\). Substituting this into the acceleration equation:
\(\text{a}(\text{t}) = -(2\pi\text{f})^2 \text{x}(\text{t}) = -4\pi^2\text{f}^2 \text{x}(\text{t})\)
This equation shows that the instantaneous acceleration at any displacement \(\text{x}\) is proportional to the square of the frequency (\(\text{f}^2\)) and the displacement (\(\text{x}\)). However, the frequency (\(\text{f}\)) itself is a fundamental parameter determining the rate of oscillation and thus the scale of acceleration. If you change the system to oscillate at a different frequency, the acceleration values will change accordingly, depending on the square of the frequency.
Analysis of the Given Statements
Let's evaluate each statement regarding the acceleration of the oscillating particle:
Statement 1: It is always in the opposite direction to the velocity
This statement is incorrect. Acceleration is always directed towards the equilibrium position (opposite to displacement, \(\text{x}\)). Velocity changes direction during the cycle. Consider a particle moving from the equilibrium position towards an extreme. Both velocity and displacement are in the same direction, but acceleration is opposite to displacement, i.e., towards the equilibrium, which is opposite to the velocity direction in this phase. However, consider a particle moving from an extreme towards the equilibrium. Velocity is towards the equilibrium, and acceleration is also towards the equilibrium. In this phase, velocity and acceleration are in the same direction. Therefore, acceleration is not always in the opposite direction to velocity.
Statement 2: It is proportional to the frequency of oscillation
This statement is the most accurate among the given options, although the relationship is technically with the square of the frequency (\(f^2\)) as shown by the equation \(\text{a} = -4\pi^2\text{f}^2 \text{x}\). The maximum magnitude of acceleration occurs at the extremes of displacement (\(|\text{x}| = \text{A}\)), given by \(|\text{a}_{\text{max}}| = 4\pi^2\text{f}^2 \text{A}\). This clearly shows that the magnitude of the acceleration is dependent on and proportional to the square of the frequency. In the context of multiple-choice questions, "proportional to the frequency" often implies that the quantity depends on the frequency, and changing the frequency changes the acceleration. Thus, frequency is a key parameter determining the magnitude of acceleration in SHM.
Statement 3: It is maximum when the speed is maximum
This statement is incorrect. Speed is maximum at the equilibrium position (\(\text{x} = 0\)), where \(\text{v}_{\text{max}} = \text{A}\omega\). At \(\text{x} = 0\), the acceleration is \(\text{a} = -\omega^2(0) = 0\). Acceleration is maximum in magnitude at the extreme positions (\(\text{x} = \pm \text{A}\)), where \(\text{v} = 0\). Thus, acceleration is minimum (zero) when speed is maximum, and acceleration is maximum when speed is minimum (zero).
Statement 4: It decreases as potential energy increases
This statement is incorrect. Potential energy (\(\text{PE}\)) in SHM for a spring-mass system is given by \(\text{PE} = \frac{1}{2}\text{kx}^2\), where \(\text{k}\) is the spring constant. Since \(\omega^2 = \text{k}/\text{m}\), \(\text{k} = \text{m}\omega^2\). So \(\text{PE} = \frac{1}{2}\text{m}\omega^2\text{x}^2\). Potential energy increases as the magnitude of displacement \(|\text{x}|\) increases. The magnitude of acceleration is \(|\text{a}| = \omega^2 |\text{x}|\). As \(|\text{x}|\) increases, potential energy increases, and the magnitude of acceleration \(|\text{a}|\) also increases. Therefore, acceleration magnitude increases as potential energy increases.
Based on the analysis, Statement 2 is the only one that correctly describes a relationship involving the acceleration in SHM, even if the proportionality is with the square of the frequency.
Revision Table: Properties of SHM
| Quantity | Position (x) | Velocity (v) | Acceleration (a) |
|---|---|---|---|
| At Equilibrium (\(\text{x}=0\)) | 0 | Maximum (\(\pm \text{A}\omega\)) | Minimum (0) |
| At Extremes (\(\text{x}=\pm\text{A}\)) | Maximum (\(\pm \text{A}\)) | Minimum (0) | Maximum (\(\mp \text{A}\omega^2\)) |
| Relationship | \(\text{x}(\text{t}) = \text{A} \cos(\omega \text{t} + \phi)\) | \(\text{v} = \pm \omega \sqrt{\text{A}^2 - \text{x}^2}\) | \(\text{a} = -\omega^2 \text{x}\) |
Additional Information on SHM
Simple Harmonic Motion is an idealized model but is fundamental to understanding oscillations and waves. Any system that is displaced slightly from a stable equilibrium position will undergo SHM if the restoring force is approximately linearly proportional to the displacement.
- Energy in SHM: The total mechanical energy (sum of kinetic and potential energy) in SHM is conserved in the absence of damping forces. Total Energy = \(\frac{1}{2}\text{mv}^2 + \frac{1}{2}\text{kx}^2 = \frac{1}{2}\text{kA}^2 = \frac{1}{2}\text{m}\omega^2\text{A}^2\), which is constant.
- Time Period and Frequency: The time period (\(\text{T}\)) is the time taken for one complete oscillation. The frequency (\(\text{f}\)) is the number of oscillations per unit time (\(\text{f} = 1/\text{T}\)). Angular frequency \(\omega = 2\pi\text{f} = 2\pi/\text{T}\). These parameters depend on the properties of the oscillating system (e.g., mass and spring constant for a spring-mass system, length and gravity for a simple pendulum).
- Phase: The phase \((\omega \text{t} + \phi)\) describes the state of the oscillation at any given time, indicating the position and direction of motion.
Similar Questions
A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g/2. The time period of the second pendulum would be:
(in terms of period T of the first pendulum)
Important Questions from Simple Harmonic Motion
- The displacement of a particle is given by $y(t) = K + P \sin^2(\omega t) + Q \sin(\omega t) \cos(\omega t)$. If this represents a simple harmonic motion, the amplitude of its oscillation is:
- Which one of the following equations of motion represents simple harmonic motion?
Assume $A$, $B$, $C$, $D$, $m$, $k$, and $\omega$ are all positive constants. In simple harmonic motion, the particle velocity lags behind the displacement by a phase angle of __________.
- A particle undergoes simple harmonic motion. Determine the phase difference between its instantaneous velocity and instantaneous acceleration.
- A particle executes simple harmonic motion along a straight line. When its displacement from the mean position is $x$, its speed is $v$. If the displacement becomes $2x$, its speed reduces to $v/2$. What is the amplitude of the oscillation?
