The displacement of a particle along the x-axis is given by `x = a sin^(2) omega t`. The motion of the particle corresponds to

Options

A. simple harmonic motion of frequency `omega//pi`
B. simple harmonic m otion of frequency`3omega//2pi`
C. non simple harmonic motion
D. simple harmonic motion of frequency `omega//2pi`

Answer

To determine whether the motion of the particle described by the displacement equation \( x = a \sin^2(\omega t) \) corresponds to simple harmonic motion (SHM) or non-simple harmonic motion, we can follow these steps:

### Step 1: Analyze the given displacement equation
The displacement of the particle is given by:
\[ x = a \sin^2(\omega t) \]

### Step 2: Rewrite the sine squared term
We can use the trigonometric identity:
\[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]
Applying this identity, we rewrite the displacement:
\[ x = a \sin^2(\omega t) = a \cdot \frac{1 - \cos(2\omega t)}{2} = \frac{a}{2} - \frac{a}{2} \cos(2\omega t) \]

### Step 3: Identify the form of the equation
The equation can be expressed as:
\[ x = \frac{a}{2} - \frac{a}{2} \cos(2\omega t) \]
This indicates that the motion is not centered around the origin (the average position is \(\frac{a}{2}\)), and it does not have the standard form of SHM, which is typically \(x = A \cos(\omega t + \phi)\).

### Step 4: Determine the velocity
To check if the motion is SHM, we calculate the velocity \(v\) by differentiating \(x\) with respect to time \(t\):
\[ v = \frac{dx}{dt} = -\frac{a}{2} \cdot (-\sin(2\omega t) \cdot 2\omega) = a\omega \sin(2\omega t) \]

### Step 5: Determine the acceleration
Next, we calculate the acceleration \(a\) by differentiating \(v\) with respect to time \(t\):
\[ a = \frac{dv}{dt} = a\omega \cdot 2\omega \cos(2\omega t) = 2a\omega^2 \cos(2\omega t) \]

### Step 6: Check the relationship between acceleration and displacement
For SHM, the acceleration should be directly proportional to the negative displacement:
\[ a \propto -x \]
In our case, we have:
\[ a = 2a\omega^2 \cos(2\omega t) \]
And since \(x = \frac{a}{2} - \frac{a}{2} \cos(2\omega t)\), we see that the acceleration is not proportional to \(-x\).

### Conclusion
Since the acceleration is not directly proportional to the negative of the displacement, the motion described by the equation \(x = a \sin^2(\omega t)\) is **non-simple harmonic motion**.

### Final Answer
The motion of the particle corresponds to **non-simple harmonic motion**.

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