Which one of the following four particles, whose displacement x and acceleration a xare related as following, is executing simple harmonic motion?
a x= -3x 2
Understanding Simple Harmonic Motion (SHM)
Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts towards the equilibrium position. A key characteristic of SHM is the relationship between the acceleration and the displacement of the particle.
Mathematically, the condition for a particle to be executing SHM is that its acceleration \(a_x\) along the direction of motion is directly proportional to its displacement \(x\) from the equilibrium position (usually taken as \(x=0\)) and is always directed towards the equilibrium position. This is represented by the equation:
\[ a_x = -\omega^2 x \]
Here, \(a_x\) is the acceleration, \(x\) is the displacement, and \(\omega\) (omega) is a positive constant called the angular frequency. The negative sign is crucial; it indicates that the acceleration is always in the opposite direction to the displacement, pulling the particle back towards the equilibrium point.
Analyzing the Given Relationships for SHM
Let's examine each of the given relationships between displacement \(x\) and acceleration \(a_x\) to see which one fits the condition \(a_x = -\omega^2 x\).
Option 1: \(a_x = +3x\)
- In this case, the acceleration is proportional to the displacement, but the constant of proportionality is positive (+3).
- This means when \(x\) is positive, \(a_x\) is positive (away from equilibrium). When \(x\) is negative, \(a_x\) is negative (away from equilibrium).
- The acceleration is not directed towards the equilibrium position (\(x=0\)). Therefore, this is not SHM.
Option 2: \(a_x = +3x^2\)
- Here, the acceleration is proportional to the square of the displacement (\(x^2\)), not the displacement \(x\) itself. The relationship is not linear with respect to displacement.
- Also, \(x^2\) is always positive (or zero). This means \(a_x = +3x^2\) is always positive (or zero), regardless of whether \(x\) is positive or negative. The acceleration is always in the positive direction, not towards the equilibrium position.
- Therefore, this is not SHM.
Option 3: \(a_x = -3x^2\)
- In this relationship, the acceleration is proportional to the square of the displacement (\(x^2\)), not the displacement \(x\). The relationship is not linear with respect to displacement.
- Also, since \(x^2\) is always positive (or zero), \(a_x = -3x^2\) is always negative (or zero), regardless of whether \(x\) is positive or negative. The acceleration is always in the negative direction, not towards the equilibrium position (unless \(x=0\)).
- Therefore, this is not SHM.
Option 4: \(a_x = -3x\)
- In this case, the acceleration \(a_x\) is directly proportional to the displacement \(x\).
- The constant of proportionality is negative (-3). This matches the required form \(a_x = -\omega^2 x\), where \(\omega^2 = 3\).
- The negative sign correctly indicates that the acceleration is always directed towards the equilibrium position (\(x=0\)), opposing the displacement. For example, if \(x > 0\), \(a_x < 0\) (acceleration is towards negative x); if \(x < 0\), \(a_x > 0\) (acceleration is towards positive x).
- This relationship perfectly satisfies the condition for Simple Harmonic Motion.
Identifying the Simple Harmonic Motion Relationship
Based on the analysis, the relationship that describes Simple Harmonic Motion among the given options is the one where acceleration is directly proportional to displacement and negatively signed, indicating it's directed towards the equilibrium. This is found in Option 4: \(a_x = -3x\).
Revision Table: SHM Condition Summary
| Property | Condition for SHM | Example from Options |
|---|---|---|
| Acceleration \(a_x\) and Displacement \(x\) Relationship | \(a_x \propto -x\) | \(a_x = -3x\) |
| Mathematical Form | \(a_x = -\omega^2 x\) (where \(\omega^2\) is a positive constant) | \(a_x = -3x\) (Here \(\omega^2 = 3\)) |
| Direction of Acceleration | Always directed towards the equilibrium position (\(x=0\)), opposite to the displacement. | If \(x > 0\), \(a_x < 0\); If \(x < 0\), \(a_x > 0\). |
Additional Information on SHM and Forces
The relationship \(a_x = -\omega^2 x\) is directly related to the restoring force in SHM. According to Newton's second law, \(F_x = m a_x\). Substituting the SHM acceleration condition, we get:
\[ F_x = m (-\omega^2 x) = -m\omega^2 x \]
Since \(m\) (mass) and \(\omega^2\) are positive constants, let \(k = m\omega^2\). Then the force becomes:
\[ F_x = -kx \]
This is Hooke's Law, which describes the restoring force exerted by an ideal spring. The constant \(k\) is the spring constant, and the negative sign shows the force is always directed towards the equilibrium position (\(x=0\)), opposite to the displacement. Systems that obey this force law, such as a mass on a spring (neglecting friction), execute Simple Harmonic Motion.
Therefore, a particle executes SHM if and only if the net force acting on it is a linear restoring force proportional to its displacement from equilibrium.
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?
