A simple harmonic motion is represented by `x(t) = sin^2 omegat - 2 cos^(2) omegat`. The angular frequency of oscillation is given by

Options

A. `omega`
B. `2 omega`
C. `4 omega`
D. `(omega)/2`

Answer

To find the angular frequency of the given simple harmonic motion represented by the equation \( x(t) = \sin^2(\omega t) - 2\cos^2(\omega t) \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ x(t) = \sin^2(\omega t) - 2\cos^2(\omega t) \]

### Step 2: Use the Pythagorean identity
Recall the Pythagorean identity:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
From this, we can express \(\sin^2(\omega t)\) in terms of \(\cos^2(\omega t)\):
\[ \sin^2(\omega t) = 1 - \cos^2(\omega t) \]
Substituting this into our equation gives:
\[ x(t) = (1 - \cos^2(\omega t)) - 2\cos^2(\omega t) \]

### Step 3: Simplify the equation
Now, simplify the equation:
\[ x(t) = 1 - \cos^2(\omega t) - 2\cos^2(\omega t) = 1 - 3\cos^2(\omega t) \]

### Step 4: Use the double angle identity
Next, we can use the double angle identity for cosine:
\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]
Applying this to our equation:
\[ x(t) = 1 - 3\left(\frac{1 + \cos(2\omega t)}{2}\right) \]
This simplifies to:
\[ x(t) = 1 - \frac{3}{2} - \frac{3}{2}\cos(2\omega t) = -\frac{1}{2} - \frac{3}{2}\cos(2\omega t) \]

### Step 5: Identify the angular frequency
The equation can be rewritten as:
\[ x(t) = -\frac{1}{2} - \frac{3}{2}\cos(2\omega t) \]
In this form, we can see that the term \(\cos(2\omega t)\) indicates that the angular frequency of the oscillation is \(2\omega\).

### Conclusion
Thus, the angular frequency of oscillation is:
\[ \text{Angular frequency} = 2\omega \]