The displacement of a particle is represented by the equation `y=sin^(3)omegat`. The motion is

Options

A. non-periodic
B. periodic but not simple harmonic
C. simple harmonic with period `2pi//omega`
D. simple harmonic with period `pi//omega`

Explanation

To determine the nature of the motion represented by the displacement equation \( y = \sin^3(\omega t) \), we will analyze the equation step by step.

### Step 1: Identify the form of the displacement equation
The given displacement equation is:
\[ y = \sin^3(\omega t) \]

### Step 2: Use trigonometric identities
We can rewrite \( \sin^3(\theta) \) using a trigonometric identity:
\[ \sin^3(\theta) = \frac{3\sin(\theta) - \sin(3\theta)}{4} \]
Applying this identity, we have:
\[ y = \frac{1}{4} (3\sin(\omega t) - \sin(3\omega t)) \]

### Step 3: Determine periodicity
Both \( \sin(\omega t) \) and \( \sin(3\omega t) \) are periodic functions. The period of \( \sin(\omega t) \) is \( T_1 = \frac{2\pi}{\omega} \) and the period of \( \sin(3\omega t) \) is \( T_2 = \frac{2\pi}{3\omega} \). The least common multiple of these two periods will determine the period of the combined function.

Since \( \sin(\omega t) \) has a longer period, the overall function \( y \) will also be periodic. Therefore, we conclude that the motion is periodic.

### Step 4: Check for simple harmonic motion (SHM)
For a motion to be classified as simple harmonic motion, the acceleration must be directly proportional to the displacement and directed towards the mean position. We can find the acceleration by differentiating the displacement function twice.

1. First derivative (velocity):
\[ \frac{dy}{dt} = \frac{1}{4} (3\omega \cos(\omega t) - 3\omega \cos(3\omega t)) \]

2. Second derivative (acceleration):
\[ \frac{d^2y}{dt^2} = \frac{1}{4} (-3\omega^2 \sin(\omega t) + 9\omega^2 \sin(3\omega t)) \]

### Step 5: Analyze the relationship between acceleration and displacement
For simple harmonic motion, we need:
\[ \frac{d^2y}{dt^2} = -k y \]
where \( k \) is a constant. In our case, the second derivative shows that the acceleration is not directly proportional to \( -y \) because it involves both \( \sin(\omega t) \) and \( \sin(3\omega t) \). Therefore, the motion is not simple harmonic.

### Conclusion
The motion represented by the equation \( y = \sin^3(\omega t) \) is periodic but not simple harmonic. Thus, the correct answer is:
**Periodic but not simple harmonic motion.**