A body oscillates with SHM according to the equation (in SHM unit ), `x=5"cos"(2pit+(pi)/(4))` . Its instantaneous displacement at t=1 s is

Options

A. `(sqrt(2))/(5)`m
B. `(1)/(sqrt(3))`
C. `(5)/(sqrt(2))`
D. `(1)/(2)` m

Explanation

To find the instantaneous displacement of a body oscillating in simple harmonic motion (SHM) given by the equation \( x = 5 \cos(2\pi t + \frac{\pi}{4}) \) at \( t = 1 \) second, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Equation**: The equation of motion is given as:
\[ x = 5 \cos(2\pi t + \frac{\pi}{4}) \]

2. **Substitute the Time**: We need to find the displacement at \( t = 1 \) second. Substitute \( t = 1 \) into the equation:
\[ x(1) = 5 \cos(2\pi(1) + \frac{\pi}{4}) \]

3. **Calculate the Argument of Cosine**: Simplify the argument of the cosine function:
\[ 2\pi(1) + \frac{\pi}{4} = 2\pi + \frac{\pi}{4} \]

4. **Use the Periodicity of Cosine**: Since cosine is periodic with a period of \( 2\pi \), we can simplify:
\[ \cos(2\pi + \frac{\pi}{4}) = \cos(\frac{\pi}{4}) \]

5. **Evaluate Cosine**: We know that:
\[ \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \]

6. **Substitute Back into the Equation**: Now substitute this value back into the equation for \( x \):
\[ x(1) = 5 \cdot \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} \]

7. **Final Result**: Therefore, the instantaneous displacement at \( t = 1 \) second is:
\[ x(1) = \frac{5}{\sqrt{2}} \text{ meters} \]

### Conclusion:
The correct answer is \( \frac{5}{\sqrt{2}} \) meters.