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)`

Explanation

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

### Step 1: Understand the parameters of SHM
- The small mass executes linear SHM about point O.
- The amplitude of the motion is given as \( A \).
- The time period of the oscillation is given as \( T \).

### Step 2: Write the equation for displacement in SHM
The displacement \( x \) of a mass executing SHM can be expressed using the formula:
\[ x = A \sin(\omega t) \]
where \( \omega \) is the angular frequency.

### Step 3: Calculate the angular frequency
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]

### Step 4: Substitute \( t = \frac{T}{8} \) into the displacement equation
We want to find the displacement at time \( t = \frac{T}{8} \):
\[ x = A \sin\left(\omega \cdot \frac{T}{8}\right) \]
Substituting \( \omega \):
\[ x = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{8}\right) \]

### Step 5: Simplify the expression
This simplifies to:
\[ x = A \sin\left(\frac{2\pi}{8}\right) = A \sin\left(\frac{\pi}{4}\right) \]

### Step 6: Calculate \( \sin\left(\frac{\pi}{4}\right) \)
We know that:
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus, substituting this back into the equation for \( x \):
\[ x = A \cdot \frac{1}{\sqrt{2}} = \frac{A}{\sqrt{2}} \]

### Final Result
The displacement from point O at time \( \frac{T}{8} \) is:
\[ x = \frac{A}{\sqrt{2}} \]