The displacement of a particle in S.H.M. is given by `x=5["cos" pi t + "sin" pi t]` where x is in metre. The amplitude of motion of the particle is given by

Options

A. 8 m
B. 10 m
C. `5sqrt(2)m`
D. `10sqrt(2) m`

Explanation

To find the amplitude of the motion of the particle given the displacement equation in simple harmonic motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Displacement Equation**:
The displacement of the particle is given by:
\[ x = 5 \cos(\pi t) + \sin(\pi t) \]

2. **Rewrite the Equation**:
We can express the second term with a coefficient of 5 for consistency:
\[ x = 5 \cos(\pi t) + 5 \cdot \frac{1}{5} \sin(\pi t) = 5 \cos(\pi t) + 5 \sin(\pi t) \]

3. **Recognize the Form of SHM**:
The displacement can be viewed as a combination of two SHMs:
- \( A_1 = 5 \) for \( \cos(\pi t) \)
- \( A_2 = 5 \) for \( \sin(\pi t) \)

4. **Determine the Phase Difference**:
The phase difference between \( \cos \) and \( \sin \) functions is \( \frac{\pi}{2} \) radians.

5. **Calculate the Resultant Amplitude**:
The resultant amplitude \( A_R \) can be calculated using the formula for the resultant of two perpendicular SHMs:
\[ A_R = \sqrt{A_1^2 + A_2^2} \]
Substituting the values:
\[ A_R = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} \]

6. **Simplify the Result**:
\[ A_R = \sqrt{50} = 5\sqrt{2} \]

### Final Answer:
The amplitude of motion of the particle is:
\[ \boxed{5\sqrt{2} \text{ m}} \]