A particle moves on the X-axis according to the equation `x=x_0 sin^2omegat`. The motion simple harmonic

Options

A. with amplitude `x_0`
B. with amplitude `2x_0`
C. with time period `(2pi)/omega`
D. with time period `pi/omega`

Answer

To determine whether the motion described by the equation \( x = x_0 \sin^2(\omega t) \) is simple harmonic motion (SHM), we can follow these steps:

### Step 1: Rewrite the Equation
The given equation is:
\[ x = x_0 \sin^2(\omega t) \]
We can use the trigonometric identity for sine squared:
\[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]
Applying this identity, we can rewrite \( \sin^2(\omega t) \):
\[ \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} \]
Thus, substituting this into the equation gives:
\[ x = x_0 \left(\frac{1 - \cos(2\omega t)}{2}\right) = \frac{x_0}{2} - \frac{x_0}{2} \cos(2\omega t) \]

### Step 2: Identify the Form of the Equation
Now, we can rearrange the equation:
\[ x - \frac{x_0}{2} = -\frac{x_0}{2} \cos(2\omega t) \]
This can be expressed in the standard form of SHM:
\[ x = A \cos(\omega' t) + C \]
where \( A = -\frac{x_0}{2} \) and \( C = \frac{x_0}{2} \).

### Step 3: Determine the Amplitude and Angular Frequency
From the equation \( x = \frac{x_0}{2} - \frac{x_0}{2} \cos(2\omega t) \):
- The amplitude \( A \) of the motion is given by:
\[ A = \frac{x_0}{2} \]
- The angular frequency \( \omega' \) is:
\[ \omega' = 2\omega \]

### Step 4: Conclusion
Since the motion can be expressed in the form of \( x = A \cos(\omega' t) + C \), where \( A \) is the amplitude and \( \omega' \) is the angular frequency, we conclude that the motion described by the equation \( x = x_0 \sin^2(\omega t) \) is indeed simple harmonic motion.

### Summary
- The motion is SHM because it can be expressed in the standard form of SHM.
- The amplitude of the motion is \( \frac{x_0}{2} \).
- The angular frequency is \( 2\omega \).