The displacement of a particle from its mean position (in mean is given by `y = 0.2 sin(10pi t + 1.5 pi) cos (10 pi t+ 1.5 pi)`. The motion but not `S.H.M.`

Options

A. Periodic but not `S.H.M`
B. None-periodic
C. Simple harmonic motion with period `0.1 s`
D. Simple harmonic motion with period `0.2 s`

Explanation

To determine if the motion described by the equation \( y = 0.2 \sin(10\pi t + 1.5\pi) \cos(10\pi t + 1.5\pi) \) is simple harmonic motion (SHM), we can simplify the equation using trigonometric identities.

### Step-by-Step Solution:

1. **Identify the given equation**:
\[ y = 0.2 \sin(10\pi t + 1.5\pi) \cos(10\pi t + 1.5\pi) \]

2. **Use the trigonometric identity**:
We can use the identity \( \sin A \cos A = \frac{1}{2} \sin(2A) \). Here, let \( A = 10\pi t + 1.5\pi \).

3. **Apply the identity**:
\[ y = 0.2 \cdot \frac{1}{2} \sin(2(10\pi t + 1.5\pi)) \]
\[ y = 0.1 \sin(20\pi t + 3\pi) \]

4. **Rewrite the equation**:
The equation can be rewritten as:
\[ y = 0.1 \sin(20\pi t + 3\pi) \]

5. **Identify the parameters**:
From the standard form of SHM, \( y = A \sin(\omega t + \phi) \), we can identify:
- Amplitude \( A = 0.1 \)
- Angular frequency \( \omega = 20\pi \)
- Phase constant \( \phi = 3\pi \)

6. **Calculate the period**:
The angular frequency \( \omega \) is related to the period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Therefore,
\[ T = \frac{2\pi}{20\pi} = \frac{1}{10} = 0.1 \text{ seconds} \]

7. **Conclusion**:
The equation \( y = 0.1 \sin(20\pi t + 3\pi) \) is indeed in the form of simple harmonic motion. Therefore, the original motion described by \( y = 0.2 \sin(10\pi t + 1.5\pi) \cos(10\pi t + 1.5\pi) \) is **not** simple harmonic motion because it involves a product of sine and cosine functions, which is not a standard form of SHM.