Out of the following functions representing motion of a particle which represents SHM
I. `y = sin omega t - cos omega t`
II. `y = sin^(3)omega t`
III. `y = 5 cos ((3 pi)/(4)-3 omega t)`
IV. `y = 1 + omega t + omega^(2)t^(2)`

Options

A. Only IV does not represent SHM
B. I and III
C. I and II
D. Only I

Explanation

To determine which of the given functions represents Simple Harmonic Motion (SHM), we need to analyze each function based on the criteria for SHM. The key criterion is that the acceleration of the particle must be directly proportional to the negative of its displacement from the mean position, which can be mathematically expressed as:

\[ \frac{d^2y}{dt^2} = -\omega^2 y \]

Let's evaluate each function step by step.

### Step 1: Analyze the first function \( y = \sin(\omega t) - \cos(\omega t) \)

1. **First Derivative**:
\[ \frac{dy}{dt} = \omega \cos(\omega t) + \omega \sin(\omega t) \]

2. **Second Derivative**:
\[ \frac{d^2y}{dt^2} = -\omega^2 \sin(\omega t) + \omega^2 \cos(\omega t) \]
This can be rewritten as:
\[ \frac{d^2y}{dt^2} = -\omega^2 (\sin(\omega t) - \cos(\omega t)) = -\omega^2 y \]
Since this satisfies the SHM condition, **this function represents SHM**.

### Step 2: Analyze the second function \( y = \sin^3(\omega t) \)

1. **First Derivative**:
Using the chain rule:
\[ \frac{dy}{dt} = 3 \sin^2(\omega t) \cdot \omega \cos(\omega t) \]

2. **Second Derivative**:
This will involve product and chain rules, and it will not yield a form that is proportional to \(-\sin^3(\omega t)\). Therefore, it does not satisfy the SHM condition. **This function does not represent SHM**.

### Step 3: Analyze the third function \( y = 5 \cos\left(\frac{3\pi}{4} - 3\omega t\right) \)

1. **First Derivative**:
\[ \frac{dy}{dt} = -15\omega \sin\left(\frac{3\pi}{4} - 3\omega t\right) \]

2. **Second Derivative**:
\[ \frac{d^2y}{dt^2} = -15\omega \cdot (-3\omega) \cos\left(\frac{3\pi}{4} - 3\omega t\right) = 45\omega^2 \cos\left(\frac{3\pi}{4} - 3\omega t\right) \]
This can be rewritten as:
\[ \frac{d^2y}{dt^2} = -\omega^2 y \]
Since this satisfies the SHM condition, **this function represents SHM**.

### Step 4: Analyze the fourth function \( y = 1 + \omega t + \omega^2 t^2 \)

1. **First Derivative**:
\[ \frac{dy}{dt} = \omega + 2\omega^2 t \]

2. **Second Derivative**:
\[ \frac{d^2y}{dt^2} = 2\omega^2 \]
This is a constant and does not depend on \(y\). Therefore, it does not satisfy the SHM condition. **This function does not represent SHM**.

### Conclusion:
The functions that represent SHM are:
- I. \( y = \sin(\omega t) - \cos(\omega t) \)
- III. \( y = 5 \cos\left(\frac{3\pi}{4} - 3\omega t\right) \)