A particle executes simple harmonic motion between `x=-A and x=+A`. The time taken for it to go from 0 to A/2. is `T_(1)` and to go from A/2 to A is `T_(2)`. Then:

Options

A. `T_(1)ltT_(2)`
B. `T_(1)gtT_(2)`
C. `T_(1)=T_(2)`
D. `T_(1)=2T_(2)`

Explanation

To solve the problem of comparing the time taken for a particle executing simple harmonic motion (SHM) to travel from \(0\) to \(\frac{A}{2}\) (denoted as \(T_1\)) and from \(\frac{A}{2}\) to \(A\) (denoted as \(T_2\)), we can follow these steps:

### Step 1: Understand the Motion
The particle is oscillating between \(x = -A\) and \(x = +A\). The mean position is at \(x = 0\). The particle's displacement can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant.

### Step 2: Determine the Time \(T_1\)
To find \(T_1\), we need to determine the time taken for the particle to move from \(x = 0\) to \(x = \frac{A}{2}\).

1. Set \(x = \frac{A}{2}\):
\[ \frac{A}{2} = A \sin(\omega T_1 + \phi) \]
Simplifying gives:
\[ \sin(\omega T_1 + \phi) = \frac{1}{2} \]
2. The angle whose sine is \(\frac{1}{2}\) is \(\frac{\pi}{6}\) (30 degrees). Therefore:
\[ \omega T_1 + \phi = \frac{\pi}{6} \]
3. Assuming the particle starts from the mean position (\(x = 0\)), we can take \(\phi = 0\):
\[ \omega T_1 = \frac{\pi}{6} \implies T_1 = \frac{\pi}{6\omega} \]

### Step 3: Determine the Time \(T_2\)
Next, we find \(T_2\), the time taken for the particle to move from \(x = \frac{A}{2}\) to \(x = A\).

1. Set \(x = A\):
\[ A = A \sin(\omega T_2 + \phi) \]
Simplifying gives:
\[ \sin(\omega T_2 + \phi) = 1 \]
2. The angle whose sine is \(1\) is \(\frac{\pi}{2}\) (90 degrees). Therefore:
\[ \omega T_2 + \phi = \frac{\pi}{2} \]
3. Again, assuming \(\phi = 0\):
\[ \omega T_2 = \frac{\pi}{2} \implies T_2 = \frac{\pi}{2\omega} \]

### Step 4: Compare \(T_1\) and \(T_2\)
Now we compare \(T_1\) and \(T_2\):
- \(T_1 = \frac{\pi}{6\omega}\)
- \(T_2 = \frac{\pi}{2\omega}\)

To compare:
\[ T_2 = \frac{\pi}{2\omega} = 3 \cdot \frac{\pi}{6\omega} = 3T_1 \]

### Conclusion
Thus, we find that \(T_2\) is greater than \(T_1\):
\[ T_2 > T_1 \]