Position of a particle varies as `y = cos^(2) omega t - sin^(2) omega t`. It is

Options

A. harmonic but not SHM
B. SHM with period `((pi)/(omega))`
C. SHM with period `((2pi)/(omega))`
D. periodic with period `((pi)/(omega))` but not SHM.

Answer

To solve the problem, we need to analyze the given function for the position of a particle, which is expressed as:

\[ y = \cos^2(\omega t) - \sin^2(\omega t) \]

### Step 1: Use Trigonometric Identity
We can simplify the expression using the trigonometric identity:

\[ \cos^2 A - \sin^2 A = \cos(2A) \]

In this case, let \( A = \omega t \). Therefore, we can rewrite the equation as:

\[ y = \cos(2\omega t) \]

### Step 2: Identify the Form of the Function
The function \( y = \cos(2\omega t) \) is a cosine function, which is a standard form of simple harmonic motion (SHM).

### Step 3: Determine the Angular Frequency
From the expression \( y = \cos(2\omega t) \), we can identify the angular frequency \( \omega_0 \):

\[ \omega_0 = 2\omega \]

### Step 4: Calculate the Time Period
The time period \( T \) of SHM is related to the angular frequency by the formula:

\[ T = \frac{2\pi}{\omega_0} \]

Substituting the value of \( \omega_0 \):

\[ T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \]

### Conclusion
The function \( y = \cos^2(\omega t) - \sin^2(\omega t) \) represents simple harmonic motion (SHM) with a time period of:

\[ T = \frac{\pi}{\omega} \]

Thus, the answer is that the motion is SHM with a period of \( \frac{\pi}{\omega} \).

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