A small mass executes linear `SHM` about `O` with amplitude a and period `T`. Its displacement from `O` at time `T//8` after passing through `O` is:

Options

A. `A/8`
B. `A/(2sqrt2)`
C. `A/2`
D. `A/sqrt2`

Answer

To solve the problem step by step, we will use the concepts of Simple Harmonic Motion (SHM).

### Step-by-Step Solution:

1. **Understanding SHM**:
A mass executing simple harmonic motion can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( x(t) \) is the displacement at time \( t \),
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

2. **Setting Initial Conditions**:
Since the mass passes through the origin at \( t = 0 \), we can set the phase constant \( \phi = 0 \). Thus, the equation simplifies to:
\[ x(t) = A \sin(\omega t) \]

3. **Finding Angular Frequency**:
The angular frequency \( \omega \) is related to the period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]

4. **Substituting for Displacement**:
Now, substituting \( \omega \) into the displacement equation, we have:
\[ x(t) = A \sin\left(\frac{2\pi}{T} t\right) \]

5. **Calculating Displacement at \( t = \frac{T}{8} \)**:
We need to find the displacement at \( t = \frac{T}{8} \):
\[ x\left(\frac{T}{8}\right) = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{8}\right) \]
Simplifying this gives:
\[ x\left(\frac{T}{8}\right) = A \sin\left(\frac{2\pi}{8}\right) = A \sin\left(\frac{\pi}{4}\right) \]

6. **Using the Value of \( \sin\left(\frac{\pi}{4}\right) \)**:
We know that:
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Therefore:
\[ x\left(\frac{T}{8}\right) = A \cdot \frac{1}{\sqrt{2}} = \frac{A}{\sqrt{2}} \]

7. **Final Answer**:
The displacement from the origin at time \( \frac{T}{8} \) after passing through \( O \) is:
\[ x = \frac{A}{\sqrt{2}} \]