A simple harmonic motion is represented by
`x(t) = sin^(2)omega t - 2 cos^(2) omega t`. 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 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
Start with the given equation:
\[ x(t) = \sin^2(\omega t) - 2\cos^2(\omega t) \]

### Step 2: Use Trigonometric Identities
We can use the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) to express \( \sin^2(\omega t) \) in terms of \( \cos^2(\omega t) \):
\[ \sin^2(\omega t) = 1 - \cos^2(\omega t) \]
Substituting this into the equation gives:
\[ x(t) = (1 - \cos^2(\omega t)) - 2\cos^2(\omega t) \]
\[ x(t) = 1 - 3\cos^2(\omega t) \]

### Step 3: Express in Terms of Cosine
Now, we can express \( x(t) \) in a more recognizable form:
\[ x(t) = 1 - 3\cos^2(\omega t) \]

### Step 4: Use the Double Angle Formula
Using the double angle formula \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \):
\[ x(t) = 1 - 3\left(\frac{1 + \cos(2\omega t)}{2}\right) \]
\[ x(t) = 1 - \frac{3}{2} - \frac{3}{2}\cos(2\omega t) \]
\[ x(t) = -\frac{1}{2} - \frac{3}{2}\cos(2\omega t) \]

### Step 5: Identify the Angular Frequency
The equation can be rewritten as:
\[ x(t) = -\frac{3}{2}\cos(2\omega t) - \frac{1}{2} \]
This is in the form of \( A \cos(\omega' t) + C \), where \( \omega' = 2\omega \).

Thus, the angular frequency of oscillation is:
\[ \omega' = 2\omega \]

### Final Answer
The angular frequency of oscillation is \( 2\omega \).

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