A particle undergoing SHM has the equation `x=A"sin"(2omegat+phi)` , where x represents the displacement of the particle. The kinetic energy oscillates with time period

Options

A. `(2pi)/(omega)`
B. `(pi)/(omega)`
C. `(4pi)/(omega)`
D. None of these

Explanation

To solve the problem, we need to determine the time period of the kinetic energy of a particle undergoing simple harmonic motion (SHM) given its displacement equation.

### Step-by-Step Solution:

1. **Identify the given equation**:
The displacement of the particle is given by the equation:
\[ x = A \sin(2\omega t + \phi) \]
Here, \(A\) is the amplitude, \(2\omega\) is the angular frequency, and \(\phi\) is the phase constant.

2. **Determine the angular frequency**:
From the equation, we can see that the angular frequency \(\omega'\) is:
\[ \omega' = 2\omega \]

3. **Relate angular frequency to time period**:
The relationship between angular frequency and time period \(T\) is given by:
\[ \omega' = \frac{2\pi}{T} \]
Rearranging this equation gives us:
\[ T = \frac{2\pi}{\omega'} \]

4. **Substitute the value of \(\omega'\)**:
Now we substitute \(\omega' = 2\omega\) into the time period equation:
\[ T = \frac{2\pi}{2\omega} \]

5. **Simplify the expression**:
Simplifying the equation gives:
\[ T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \]

6. **Conclusion**:
The time period of the oscillation of the kinetic energy is:
\[ T = \frac{\pi}{\omega} \]
Therefore, the correct answer is option (b) \(\frac{\pi}{\omega}\).