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))`m
C. `(1)/(2)` m
D. `(5)/(sqrt(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 given equation**:
The equation of motion is given as:
\[ x = 5 \cos(2\pi t + \frac{\pi}{4}) \]

2. **Substitute the time \( t = 1 \) second into the equation**:
\[ x = 5 \cos(2\pi(1) + \frac{\pi}{4}) \]
This simplifies to:
\[ x = 5 \cos(2\pi + \frac{\pi}{4}) \]

3. **Use the periodic property of the cosine function**:
The cosine function has a period of \( 2\pi \), so:
\[ \cos(2\pi + \frac{\pi}{4}) = \cos(\frac{\pi}{4}) \]

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

5. **Substitute back into the equation**:
Now, substitute \( \cos(\frac{\pi}{4}) \) back into the equation:
\[ x = 5 \cdot \frac{1}{\sqrt{2}} \]

6. **Simplify the expression**:
\[ x = \frac{5}{\sqrt{2}} \text{ meters} \]

### Final Answer:
The instantaneous displacement at \( t = 1 \) second is:
\[ x = \frac{5}{\sqrt{2}} \text{ meters} \]