The displacement of a particle from its mean position (in metre) is given by
`y=0.2 "sin" (10 pi t+ 1.5 pi) "cos" (10 pi t+1.5 pi)`
The motion of the particle is

Options

A. periodic but not S.H.M
B. non - periodic
C. simple harmonic motion with a period 0.1 s
D. simple harmonic motion with a period 0.2 s

Answer

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

**Given:**
\[ y = 0.2 \sin(10\pi t + 1.5\pi) \cos(10\pi t + 1.5\pi) \]

### Step 1: Use Trigonometric Identity
We can use the trigonometric identity:
\[ \sin A \cos A = \frac{1}{2} \sin(2A) \]
In our case, let \( A = 10\pi t + 1.5\pi \).

### Step 2: Rewrite the Equation
Using the identity, we can rewrite the displacement equation:
\[ y = 0.2 \cdot \frac{1}{2} \sin(2(10\pi t + 1.5\pi)) \]
\[ y = 0.1 \sin(20\pi t + 3\pi) \]

### Step 3: Identify the Standard Form of SHM
The standard form of simple harmonic motion (SHM) is:
\[ y = A \sin(\omega t + \phi) \]
Where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

From our equation:
- Amplitude \( A = 0.1 \)
- Angular frequency \( \omega = 20\pi \)
- Phase constant \( \phi = 3\pi \)

### Step 4: Calculate the Time Period
The time period \( T \) of SHM is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega = 20\pi \):
\[ T = \frac{2\pi}{20\pi} = \frac{1}{10} = 0.1 \text{ seconds} \]

### Conclusion
The motion of the particle is simple harmonic motion (SHM) with a period of 0.1 seconds.

### Final Answer
The motion of the particle is **simple harmonic motion with a period of 0.1 seconds**.

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