The displacement of a particle varies with time as `x = 12 sin omega t - 16 sin^(3) 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. 144 `omega^(2)`
D. `sqrt192 omega^(2)`

Answer

To find the maximum acceleration of the particle whose displacement varies with time as \( x = 12 \sin(\omega t) - 16 \sin^3(\omega t) \), we can follow these steps:

### Step 1: Rewrite the Displacement Equation
The given displacement is:
\[ x = 12 \sin(\omega t) - 16 \sin^3(\omega t) \]
We can factor out a common term from this equation.

### Step 2: Factor the Displacement
We can factor out 4 from the equation:
\[ x = 4(3 \sin(\omega t) - 4 \sin^3(\omega t)) \]

### Step 3: Use Trigonometric Identity
We know from trigonometric identities that:
\[ \sin(3\theta) = 3 \sin(\theta) - 4 \sin^3(\theta) \]
Using this identity, we can rewrite the displacement:
\[ x = 4 \sin(3\omega t) \]

### Step 4: Identify the Amplitude
From the rewritten equation, we can see that the amplitude \( A \) of the motion is:
\[ A = 4 \, \text{cm} \]

### Step 5: Calculate Maximum Acceleration
The maximum acceleration \( a_{\text{max}} \) in simple harmonic motion (SHM) is given by the formula:
\[ a_{\text{max}} = A \omega^2 \]
Substituting the amplitude and angular frequency:
\[ a_{\text{max}} = 4 \cdot (3\omega)^2 \]

### Step 6: Simplify the Expression
Calculating \( (3\omega)^2 \):
\[ (3\omega)^2 = 9\omega^2 \]
Thus, substituting back:
\[ a_{\text{max}} = 4 \cdot 9\omega^2 = 36\omega^2 \]

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

---