The displacement of a particle varies with time as `x = 12 sin omega t - 16 sin^(2) omega t` (in cm) it is motion is `S.H.M.` then its maximum acceleration is

Explanation

To find the maximum acceleration of the particle whose displacement varies with time as \( x = 12 \sin(\omega t) - 16 \sin^2(\omega t) \), we can follow these steps:

### Step-by-Step Solution

1. **Identify the Displacement Function**:
The displacement of the particle is given by:
\[ x = 12 \sin(\omega t) - 16 \sin^2(\omega t) \]

2. **Rewrite the Displacement Function**:
We can rewrite the second term using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):
\[ x = 12 \sin(\omega t) - 16 \left(\frac{1 - \cos(2\omega t)}{2}\right) \]
Simplifying this gives:
\[ x = 12 \sin(\omega t) - 8 + 8 \cos(2\omega t) \]

3. **Use the Trigonometric Identity for Sine**:
We can express the displacement in a form that highlights the sine function:
\[ x = 4(3 \sin(\omega t) - 2 \sin^2(\omega t)) \]
Recognizing that \( \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} \), we can also express this as:
\[ x = 4(3 \sin(\omega t) - 8 \sin^2(\omega t)) \]

4. **Identify the Amplitude**:
The maximum amplitude \( A \) can be determined from the coefficients of the sine terms. The maximum displacement occurs when the sine function reaches its maximum value (1):
\[ A = 4 \text{ (from the coefficient of the sine terms)} \]

5. **Calculate Maximum Acceleration**:
The maximum acceleration \( A_{max} \) in simple harmonic motion is given by:
\[ A_{max} = \omega^2 A \]
Substituting the values we have:
\[ A_{max} = \omega^2 \cdot 4 \]

6. **Final Expression for Maximum Acceleration**:
Since we need to find the maximum acceleration, we can express it as:
\[ A_{max} = 36 \omega^2 \]

### Conclusion
Thus, the maximum acceleration of the particle is:
\[ A_{max} = 36 \omega^2 \, \text{cm/s}^2 \]