A particle executes SHM in accordance with `x=A"sin"omegat`. If `t_(1)` is the time taken by it to reach from x=0 to `x=sqrt(3)(A//2)`and `t_(2)` is the time taken by it to reach from `x=sqrt(3)//2`A to x=A , the value of `t_(1)//t_(2)` is

Options

A. 2
B. `(1)/(2)`
C. 3
D. None of the these

Explanation

To solve the problem step-by-step, we will analyze the motion of the particle executing simple harmonic motion (SHM) given by the equation \( x = A \sin(\omega t) \).

### Step 1: Determine \( t_1 \)
We need to find the time \( t_1 \) taken by the particle to move from \( x = 0 \) to \( x = \frac{\sqrt{3}}{2} A \).

1. Set \( x = \frac{\sqrt{3}}{2} A \) in the equation:
\[ \frac{\sqrt{3}}{2} A = A \sin(\omega t_1) \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \sin(\omega t_1) = \frac{\sqrt{3}}{2} \]

2. The angle whose sine is \( \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{3} \):
\[ \omega t_1 = \frac{\pi}{3} \]

3. Therefore, we can express \( t_1 \):
\[ t_1 = \frac{\pi}{3\omega} \]

### Step 2: Determine \( t_2 \)
Now, we need to find the time \( t_2 \) taken by the particle to move from \( x = \frac{\sqrt{3}}{2} A \) to \( x = A \).

1. We know that when \( x = A \), the particle reaches its maximum displacement. Let the total time for one complete cycle be \( T \):
\[ x = A \implies A = A \sin(\omega T) \]
This implies:
\[ \sin(\omega T) = 1 \implies \omega T = \frac{\pi}{2} \]
Thus,
\[ T = \frac{\pi}{2\omega} \]

2. The time \( t_2 \) can be expressed as:
\[ t_2 = T - t_1 = \frac{\pi}{2\omega} - \frac{\pi}{3\omega} \]

3. To simplify \( t_2 \):
\[ t_2 = \frac{\pi}{2\omega} - \frac{\pi}{3\omega} = \frac{\pi}{\omega} \left( \frac{3}{6} - \frac{2}{6} \right) = \frac{\pi}{\omega} \cdot \frac{1}{6} = \frac{\pi}{6\omega} \]

### Step 3: Calculate the ratio \( \frac{t_1}{t_2} \)
Now we can find the ratio \( \frac{t_1}{t_2} \):
\[ \frac{t_1}{t_2} = \frac{\frac{\pi}{3\omega}}{\frac{\pi}{6\omega}} = \frac{6}{3} = 2 \]

### Final Answer
Thus, the value of \( \frac{t_1}{t_2} \) is \( 2 \).