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

Options

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

Explanation

To analyze the motion of the particle represented by the equation \( y = 3 \cos\left(\frac{\pi}{4} - 2\omega t\right) \), we will determine if it exhibits simple harmonic motion (SHM) and find its time period.

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

### Step 2: Differentiate the displacement equation
To check for SHM, we differentiate the displacement \( y \) with respect to time \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt}[3 \cos\left(\frac{\pi}{4} - 2\omega t\right)] \]
Using the chain rule:
\[ \frac{dy}{dt} = 3 \cdot (-\sin\left(\frac{\pi}{4} - 2\omega t\right)) \cdot \frac{d}{dt}\left(\frac{\pi}{4} - 2\omega t\right) = -6\omega \sin\left(\frac{\pi}{4} - 2\omega t\right) \]

### Step 3: Differentiate again to find acceleration
Next, we differentiate \( \frac{dy}{dt} \) to find the acceleration \( \frac{d^2y}{dt^2} \):
\[ \frac{d^2y}{dt^2} = \frac{d}{dt}[-6\omega \sin\left(\frac{\pi}{4} - 2\omega t\right)] \]
Applying the chain rule again:
\[ \frac{d^2y}{dt^2} = -6\omega \cdot \cos\left(\frac{\pi}{4} - 2\omega t\right) \cdot \frac{d}{dt}\left(\frac{\pi}{4} - 2\omega t\right) = -12\omega^2 \cos\left(\frac{\pi}{4} - 2\omega t\right) \]

### Step 4: Relate acceleration to displacement
From the equation \( \cos\left(\frac{\pi}{4} - 2\omega t\right) \), we can express it in terms of \( y \):
\[ \cos\left(\frac{\pi}{4} - 2\omega t\right) = \frac{y}{3} \]
Substituting this into the acceleration equation gives:
\[ \frac{d^2y}{dt^2} = -12\omega^2 \cdot \frac{y}{3} = -4\omega^2 y \]

### Step 5: Confirm SHM condition
The equation \( \frac{d^2y}{dt^2} = -4\omega^2 y \) confirms that the motion is simple harmonic since it satisfies the condition:
\[ \frac{d^2y}{dt^2} \propto -y \]

### Step 6: Determine the time period of SHM
The angular frequency \( \omega' \) for this SHM is \( \sqrt{4\omega^2} = 2\omega \). The time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega'} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \]

### Conclusion
The motion of the particle is simple harmonic motion (SHM) with a time period of \( \frac{\pi}{\omega} \).

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