The displacement of a particle is represented by the equation `y=3cos((pi)/(4)-2omegat).`
The motion of the particle is

Options

A. simple harmonic with period `2pi//omega`
B. simple harmonic with period `pi//omega`
C. periodic but not simple harmonic
D. non-periodic

Explanation

To determine the nature of the motion of the particle described by the equation \( y = 3 \cos\left(\frac{\pi}{4} - 2\omega t\right) \), we will analyze the equation step by step.

### Step 1: Identify the form of the equation
The given equation is of the form:
\[ y = A \cos(\phi - \omega t) \]
where \( A = 3 \) (the amplitude), \( \phi = \frac{\pi}{4} \) (the phase constant), and \( \omega = 2\omega \) (the angular frequency).

### Step 2: Check for Simple Harmonic Motion (SHM)
For a motion to be classified as Simple Harmonic Motion (SHM), it must satisfy the condition:
\[ a = -\omega^2 y \]
where \( a \) is the acceleration.

### Step 3: Calculate the velocity
The velocity \( v \) is the first derivative of displacement \( y \) with respect to time \( t \):
\[ v = \frac{dy}{dt} = \frac{d}{dt}[3 \cos\left(\frac{\pi}{4} - 2\omega t\right)] \]
Using the chain rule:
\[ v = -3 \cdot 2\omega \sin\left(\frac{\pi}{4} - 2\omega t\right) = -6\omega \sin\left(\frac{\pi}{4} - 2\omega t\right) \]

### Step 4: Calculate the acceleration
The acceleration \( a \) is the derivative of velocity \( v \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}[-6\omega \sin\left(\frac{\pi}{4} - 2\omega t\right)] \]
Using the chain rule again:
\[ a = -6\omega \cdot (-2\omega) \cos\left(\frac{\pi}{4} - 2\omega t\right) = 12\omega^2 \cos\left(\frac{\pi}{4} - 2\omega t\right) \]

### Step 5: Relate acceleration to displacement
Now, we can express \( \cos\left(\frac{\pi}{4} - 2\omega t\right) \) in terms of \( y \):
\[ \cos\left(\frac{\pi}{4} - 2\omega t\right) = \frac{y}{3} \]
Substituting this back into the acceleration equation:
\[ a = 12\omega^2 \left(\frac{y}{3}\right) = 4\omega^2 y \]
This shows that:
\[ a = -4\omega^2 y \]
which satisfies the condition for SHM.

### Step 6: Determine the period of motion
The period \( T \) of SHM is given by:
\[ T = \frac{2\pi}{\omega} \]
Here, since \( \omega = 2\omega \), we substitute:
\[ T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \]

### Conclusion
The motion of the particle is Simple Harmonic Motion (SHM) with a period of \( \frac{\pi}{\omega} \).

### Final Answer
The motion of the particle is **Simple Harmonic Motion with period \( \frac{\pi}{\omega} \)**.

---