A particle is executing simple harmonic motion with a period of T seconds and amplitude a metre . The shortest time it takes to reach a point `a/sqrt2` from its mean position in seconds is

Options

A. T
B. `(T)/(4)`
C. `(T)/(8)`
D. `(T)/(16)`

Explanation

To solve the problem step by step, we will analyze the motion of a particle executing simple harmonic motion (SHM) and find the time it takes to reach a position \( \frac{a}{\sqrt{2}} \) from its mean position.

### Step 1: Understand the parameters of SHM
The particle has:
- Amplitude \( a \) (the maximum displacement from the mean position).
- Time period \( T \) (the time taken to complete one full cycle of motion).

### Step 2: Write the equation of motion
The position \( x \) of a particle in SHM can be expressed as:
\[ x = a \sin(\omega t) \]
where \( \omega \) is the angular frequency.

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

### Step 4: Set up the equation for the given position
We want to find the time \( t \) when the particle is at the position \( x = \frac{a}{\sqrt{2}} \):
\[ \frac{a}{\sqrt{2}} = a \sin(\omega t) \]
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ \frac{1}{\sqrt{2}} = \sin(\omega t) \]

### Step 5: Solve for \( \omega t \)
The value of \( \sin(\theta) = \frac{1}{\sqrt{2}} \) corresponds to:
\[ \omega t = \frac{\pi}{4} \]
This is the first instance where the sine function takes this value in the first quadrant.

### Step 6: Substitute \( \omega \) in the equation
Now, substituting \( \omega = \frac{2\pi}{T} \) into the equation:
\[ \frac{2\pi}{T} t = \frac{\pi}{4} \]

### Step 7: Solve for \( t \)
Rearranging gives:
\[ t = \frac{\pi}{4} \cdot \frac{T}{2\pi} = \frac{T}{8} \]

### Conclusion
Thus, the shortest time it takes to reach the point \( \frac{a}{\sqrt{2}} \) from the mean position is:
\[ \boxed{\frac{T}{8}} \]