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 with 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 of motion**:
The equation given is:
\[ x = 5 \cos(2\pi t + \frac{\pi}{4}) \]

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

3. **Use the periodic property of cosine**:
The cosine function is periodic with a period of \( 2\pi \). Therefore, we can simplify \( 2\pi + \frac{\pi}{4} \) as:
\[ \cos(2\pi + \frac{\pi}{4}) = \cos(\frac{\pi}{4}) \]

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

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

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

### Conclusion:
The correct option for the instantaneous displacement at \( t = 1 \) second is \( \frac{5}{\sqrt{2}} \).

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