The displacement of a perticle 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

Options

A. `12 omega^(2)`
B. `36 omega^(2)`
C. `133 omega^(2)`
D. `sqrt(192) omega^(2)`

Explanation

To solve the problem, we need to find the maximum acceleration of a particle whose displacement varies with time as given by the equation:

\[ x = 12 \sin(\omega t) - 16 \sin^2(\omega t) \]

### Step 1: Rewrite the equation
We can use the trigonometric identity for \(\sin^2(\theta)\):
\[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]
Applying this identity to our equation:

\[ x = 12 \sin(\omega t) - 16 \left(\frac{1 - \cos(2\omega t)}{2}\right) \]
This simplifies to:
\[ x = 12 \sin(\omega t) - 8 + 8 \cos(2\omega t) \]
Rearranging gives:
\[ x = 12 \sin(\omega t) + 8 \cos(2\omega t) - 8 \]

### Step 2: Use the identity for \(\sin(3\theta)\)
We can express the equation in terms of \(\sin(3\theta)\) using the identity:
\[ \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) \]
We can factor out a common term from the original equation:
\[ x = 4(3\sin(\omega t) - 4\sin^3(\omega t)) \]
This can be rewritten as:
\[ x = 4 \sin(3\omega t) \]
Thus, we have:
\[ x = 4 \sin(3\omega t) \]

### Step 3: Identify the amplitude
From the equation \(x = 4 \sin(3\omega t)\), we can identify the amplitude \(A\):
\[ A = 4 \text{ cm} \]

### Step 4: Calculate the maximum acceleration
The formula for maximum acceleration \(a_{\text{max}}\) in simple harmonic motion is given by:
\[ a_{\text{max}} = \omega^2 A \]
Substituting the values we have:
\[ a_{\text{max}} = (3\omega)^2 \cdot 4 \]
Calculating this gives:
\[ a_{\text{max}} = 9\omega^2 \cdot 4 = 36\omega^2 \]

### Final Answer
Thus, the maximum acceleration is:
\[ \boxed{36\omega^2} \text{ cm/s}^2 \]