The x-coordinate of a particle moving on x-axis is given by `x = 3 sin 100 t + 8 cos^(2) 50 t`, where x is in cm and t is time in seconds. Which of the following is/are correct about this motion

Options

A. the motion of the particle is not S.H.M
B. the amplitude of the S.H.M of the particle is 5 cm
C. the amplitude of the resultant S.H.M is `sqrt(73) cm`
D. the maximum displacement of the particle from the origin is 9 cm

Explanation

To solve the problem, we need to analyze the motion of the particle given by the equation:

\[ x = 3 \sin(100t) + 8 \cos^2(50t) \]

### Step 1: Simplify the equation
We start by simplifying the term \( \cos^2(50t) \) using the trigonometric identity:

\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]

Applying this identity, we have:

\[ \cos^2(50t) = \frac{1 + \cos(100t)}{2} \]

Substituting this back into the equation for \( x \):

\[ x = 3 \sin(100t) + 8 \left( \frac{1 + \cos(100t)}{2} \right) \]

This simplifies to:

\[ x = 3 \sin(100t) + 4(1 + \cos(100t)) \]
\[ x = 3 \sin(100t) + 4 + 4 \cos(100t) \]
\[ x = 4 + 3 \sin(100t) + 4 \cos(100t) \]

### Step 2: Rearranging the equation
Now, we can rearrange the equation:

\[ x - 4 = 3 \sin(100t) + 4 \cos(100t) \]

### Step 3: Finding the amplitude
To express \( x - 4 \) in the standard form of simple harmonic motion (SHM), we can find the resultant amplitude of the terms \( 3 \sin(100t) + 4 \cos(100t) \).

The amplitude \( A \) of the resultant SHM can be calculated using the formula:

\[ A = \sqrt{a^2 + b^2} \]

where \( a = 3 \) and \( b = 4 \):

\[ A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 4: Maximum displacement from the origin
The maximum displacement from the origin occurs when the particle is at its maximum amplitude. Since the particle oscillates around \( x = 4 \), the maximum displacement from the origin can be calculated as:

\[ \text{Maximum displacement} = 4 + A = 4 + 5 = 9 \text{ cm} \]

### Step 5: Conclusion about the motion
Now we can summarize the findings:

1. The motion of the particle is indeed SHM since it can be expressed in the form \( x = A \sin(\omega t + \phi) + C \).
2. The amplitude of the SHM is \( 5 \) cm.
3. The maximum displacement from the origin is \( 9 \) cm.

### Final Answers
- The motion of the particle is SHM (False for the first option).
- The amplitude of SHM of the particle is \( 5 \) cm (True).
- The amplitude of resultant SHM is not \( \sqrt{73} \) cm (False).
- The maximum displacement of the particle from the origin is \( 9 \) cm (True).