A particle moves on the X-axis according to the equation `x=5sin^2omegat`. The motion simple harmonic

Options

A. with amplitude `(5//2)`
B. with amplitude `10`
C. with time period `(2pi//omega)`
D. with time period `(pi//omega)`

Explanation

To solve the problem step by step, we will analyze the given equation of motion and determine the amplitude and time period of the motion.

### Step 1: Analyze the given equation
The equation of motion is given as:
\[ x = 5 \sin^2(\omega t) \]

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

### Step 3: Substitute the identity into the equation
Substituting this back into the equation for \(x\):
\[ x = 5 \left(\frac{1 - \cos(2\omega t)}{2}\right) \]
This simplifies to:
\[ x = \frac{5}{2} (1 - \cos(2\omega t)) \]

### Step 4: Separate the equation into constant and cosine terms
Now, we can expand this equation:
\[ x = \frac{5}{2} - \frac{5}{2} \cos(2\omega t) \]

### Step 5: Identify the amplitude
In the equation \(x = A - B \cos(2\omega t)\), the amplitude of the oscillation is given by the coefficient of the cosine term. Here, the amplitude is:
\[ A = \frac{5}{2} \]

### Step 6: Determine the time period
The angular frequency \(\omega\) in the cosine term is \(2\omega\). The time period \(T\) is given by:
\[ T = \frac{2\pi}{\text{angular frequency}} \]
Thus, substituting \(2\omega\):
\[ T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \]

### Conclusion
From the analysis, we find:
- The amplitude of the motion is \(\frac{5}{2}\).
- The time period of the motion is \(\frac{\pi}{\omega}\).

### Final Answer
- Amplitude: \(\frac{5}{2}\)
- Time Period: \(\frac{\pi}{\omega}\)

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