Which of the following represents a SHM?

Options

A. `sin omega t - cos omega t`
B. `sin omega t + cos omega t`
C. `sin omega t + 2 cos omega t`
D. All of these

Answer

To determine which of the given options represents simple harmonic motion (SHM), we need to check if the equations satisfy the condition of SHM, which is given by:

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

This means that the acceleration of the system is proportional to the displacement and directed towards the mean position.

Let's analyze each option step by step.

### Step 1: Analyze Option A
Given:
\[ x(t) = \sin(\omega t) + \cos(\omega t) \]

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

2. **Second Derivative (Acceleration)**:
\[ \frac{d^2x}{dt^2} = -\omega^2 \sin(\omega t) - \omega^2 \cos(\omega t) = -\omega^2 (\sin(\omega t) + \cos(\omega t)) \]

3. **Check if it satisfies SHM**:
\[ \frac{d^2x}{dt^2} = -\omega^2 x \]
Thus, Option A represents SHM.

### Step 2: Analyze Option B
Given:
\[ x(t) = \sin(\omega t) - \cos(\omega t) \]

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

2. **Second Derivative (Acceleration)**:
\[ \frac{d^2x}{dt^2} = -\omega^2 \sin(\omega t) + \omega^2 \cos(\omega t) = -\omega^2 (\sin(\omega t) - \cos(\omega t)) \]

3. **Check if it satisfies SHM**:
\[ \frac{d^2x}{dt^2} = -\omega^2 x \]
Thus, Option B also represents SHM.

### Step 3: Analyze Option C
Given:
\[ x(t) = \sin(\omega t) + 2\cos(\omega t) \]

1. **First Derivative (Velocity)**:
\[ \frac{dx}{dt} = \omega \cos(\omega t) - 2\omega \sin(\omega t) \]

2. **Second Derivative (Acceleration)**:
\[ \frac{d^2x}{dt^2} = -\omega^2 \sin(\omega t) - 2\omega^2 \cos(\omega t) = -\omega^2 (\sin(\omega t) + 2\cos(\omega t)) \]

3. **Check if it satisfies SHM**:
\[ \frac{d^2x}{dt^2} = -\omega^2 x \]
Thus, Option C also represents SHM.

### Conclusion
Since all three options A, B, and C satisfy the condition for SHM, the correct answer is that all of these represent SHM.