A particle moves along y - axis according to the equation `y("in cm") = 3 sin 100pi t + 8sin^(2) 50pi t - 6`

Options

A. the particle perform SHM
B. the amplitude of the particle's oscillation is `5cm`
C. the mean position of the particle is at `y = - 2cm`
D. the particle does not perform SHM

Explanation

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

**Given equation:**
\[ y(t) = 3 \sin(100 \pi t) + 8 \sin^2(50 \pi t) - 6 \]

### Step 1: Simplify the Equation
First, we need to simplify the term \( \sin^2(50 \pi t) \). We can use the trigonometric identity:
\[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \]

Applying this identity:
\[ \sin^2(50 \pi t) = \frac{1 - \cos(100 \pi t)}{2} \]

Now substituting this back into the equation:
\[ y(t) = 3 \sin(100 \pi t) + 8 \left( \frac{1 - \cos(100 \pi t)}{2} \right) - 6 \]
\[ = 3 \sin(100 \pi t) + 4(1 - \cos(100 \pi t)) - 6 \]
\[ = 3 \sin(100 \pi t) + 4 - 4 \cos(100 \pi t) - 6 \]
\[ = 3 \sin(100 \pi t) - 4 \cos(100 \pi t) - 2 \]

### Step 2: Rewrite in Standard SHM Form
The equation can be rewritten in a form that resembles the standard SHM equation. The standard form is:
\[ y(t) = A \sin(\omega t + \phi) + k \]

Where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant,
- \( k \) is the mean position.

To express \( 3 \sin(100 \pi t) - 4 \cos(100 \pi t) \) in the form \( R \sin(100 \pi t + \phi) \), we can find \( R \) and \( \phi \) using:
\[ R = \sqrt{(3^2 + (-4)^2)} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Find the Phase Constant
To find the phase constant \( \phi \):
\[ \tan(\phi) = \frac{-4}{3} \]
Thus,
\[ \phi = \tan^{-1}\left(\frac{-4}{3}\right) \]

### Step 4: Identify Mean Position
The mean position \( k \) is given by the constant term in the equation:
\[ k = -2 \]

### Step 5: Conclusion
From the analysis:
1. The particle performs SHM since the equation can be expressed in the standard SHM form.
2. The amplitude of the oscillation is \( 5 \, \text{cm} \).
3. The mean position of the particle is at \( y = -2 \, \text{cm} \).

### Final Answer
Thus, the correct statements are:
- The particle performs SHM.
- The amplitude of the particle's oscillation is \( 5 \, \text{cm} \).
- The mean position of the particle is at \( y = -2 \, \text{cm} \).