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 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
We know from trigonometric identities that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Using this identity, we can express \(\sin^2(\omega t)\) in terms of \(\cos^2(\omega t)\):
\[ \sin^2(\omega t) = 1 - \cos^2(\omega t) \]

### Step 3: Substitute into the equation
Substituting \(\sin^2(\omega t)\) into the original equation gives:
\[ x(t) = (1 - \cos^2(\omega t)) - 2 \cos^2(\omega t) \]
This simplifies to:
\[ x(t) = 1 - 3 \cos^2(\omega t) \]

### Step 4: Rewrite the equation in terms of cosine
We can express \(\cos^2(\omega t)\) using the double angle formula:
\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]
Thus, we rewrite \(\cos^2(\omega t)\):
\[ x(t) = 1 - 3 \left(\frac{1 + \cos(2\omega t)}{2}\right) \]
This simplifies to:
\[ x(t) = 1 - \frac{3}{2} - \frac{3}{2} \cos(2\omega t) \]
\[ x(t) = -\frac{1}{2} - \frac{3}{2} \cos(2\omega t) \]

### Step 5: Identify the angular frequency
From the equation \(x(t) = -\frac{1}{2} - \frac{3}{2} \cos(2\omega t)\), we can see that the term \(\cos(2\omega t)\) indicates that the angular frequency of the oscillation is:
\[ \text{Angular frequency} = 2\omega \]

### Final Answer
Thus, the angular frequency of the oscillation is \(2\omega\).

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