A particle is performing SHM along x-axis such that its acceleration along x-axis is : `a = 2 -x` where a is in `m//s^(2)` and x is in meter. If speed of the particle at x = 1 is zero in INCORRECT statement is :

Options

A. time period of oscillation is `pi` second
B. amplitude of oscillation is 1 m
C. speed of the particle at `x = 2.5 m` is `(sqrt(3))/(2)` m/s
D. speed of particle at x = 3 m is zero

Explanation

To solve the problem, we need to analyze the given acceleration function and determine the characteristics of the simple harmonic motion (SHM) of the particle.

### Step 1: Understand the acceleration function
The acceleration of the particle is given by:
\[ a = 2 - x \]
This indicates that the acceleration is dependent on the position \( x \).

### Step 2: Identify the mean position
In SHM, the mean position (equilibrium position) is where the acceleration is zero. Setting the acceleration to zero:
\[ 0 = 2 - x \]
\[ x = 2 \]
Thus, the mean position is at \( x = 2 \).

### Step 3: Determine the extremes of motion
The extremes of motion occur when the particle reaches its maximum displacement from the mean position. Since the acceleration function is linear, we can find the points where the particle stops (i.e., where the velocity is zero):
- The left extreme occurs when \( x = 1 \) (as given in the problem).
- The right extreme can be found by setting the acceleration to zero again:
\[ 0 = 2 - x \]
\[ x = 2 \] (mean position)
To find the right extreme, we can also consider the symmetry of SHM. The distance from the mean position to the left extreme is \( 2 - 1 = 1 \), so the right extreme will be:
\[ x = 2 + 1 = 3 \]

### Step 4: Determine the angular frequency and time period
The angular frequency \( \omega \) is given by the coefficient of \( x \) in the acceleration function:
\[ \omega^2 = 1 \quad \Rightarrow \quad \omega = 1 \, \text{rad/s} \]
The time period \( T \) of the oscillation is given by:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{1} = 2\pi \, \text{seconds} \]

### Step 5: Calculate the maximum velocity
The maximum velocity \( v_{\text{max}} \) in SHM is given by:
\[ v_{\text{max}} = \omega A \]
Where \( A \) is the amplitude. The amplitude is the distance from the mean position to either extreme:
\[ A = 1 \, \text{meter} \]
Thus:
\[ v_{\text{max}} = 1 \times 1 = 1 \, \text{m/s} \]

### Step 6: Calculate the velocity at \( x = 1 \)
Using the formula for velocity in SHM:
\[ v = \sqrt{A^2 - x^2} \]
At \( x = 1 \):
\[ v = \sqrt{1^2 - 1^2} = \sqrt{1 - 1} = \sqrt{0} = 0 \, \text{m/s} \]
This confirms that the speed of the particle at \( x = 1 \) is indeed zero.

### Step 7: Identify the incorrect statement
Now, we need to evaluate the options given in the question to find the incorrect statement. Based on our calculations:
- The time period is \( 2\pi \) seconds, which is approximately \( 6.28 \) seconds.
- The amplitude is \( 1 \) meter.
- The maximum velocity is \( 1 \) m/s.

If one of the options states that the time period is \( \pi \) seconds, that would be the incorrect statement.

### Conclusion
The incorrect statement is that the time period of the oscillation is \( \pi \) seconds.