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, we need to find the ratio \( \frac{t_1}{t_2} \) where \( t_1 \) is the time taken to move from \( x = 0 \) to \( x = \frac{\sqrt{3}}{2}A \) and \( t_2 \) is the time taken to move from \( x = \frac{\sqrt{3}}{2}A \) to \( x = A \).

### Step 1: Determine \( t_1 \)

The position of the particle in SHM is given by:
\[ x = A \sin(\omega t) \]
For \( t_1 \), we set \( x = \frac{\sqrt{3}}{2}A \):
\[ \frac{\sqrt{3}}{2}A = A \sin(\omega t_1) \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{\sqrt{3}}{2} = \sin(\omega t_1) \]
To find \( \omega t_1 \), we take the inverse sine:
\[ \omega t_1 = \frac{\pi}{3} \]
Thus,
\[ t_1 = \frac{\pi}{3\omega} \]

### Step 2: Determine \( t_2 \)

Next, for \( t_2 \), we need to find the time taken to move from \( x = \frac{\sqrt{3}}{2}A \) to \( x = A \):
\[ A = A \sin(\omega t_2) \]
Dividing both sides by \( A \):
\[ 1 = \sin(\omega t_2) \]
This gives:
\[ \omega t_2 = \frac{\pi}{2} \]
Thus,
\[ t_2 = \frac{\pi}{2\omega} \]

### Step 3: Calculate the ratio \( \frac{t_1}{t_2} \)

Now we can find the ratio:
\[ \frac{t_1}{t_2} = \frac{\frac{\pi}{3\omega}}{\frac{\pi}{2\omega}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3} \]

### Final Answer
The value of \( \frac{t_1}{t_2} \) is:
\[ \frac{t_1}{t_2} = \frac{2}{3} \]