A particle of mass m executing SHM with amplitude A and angular frequency `omega`. The average value of the kinetic energy and potential energy over a period is

Options

A. `0,(1)/(2)momega^(2)A^(2)`
B. `(1)/(2)momega^(2)A^(2),0`
C. `(1)/(2)momega^(2)A^(2),(1)/(2)momega^(2)A^(2)`
D. `(1)/(4)momega^(2)A^(2),(1)/(4)momega^(2)A^(2)`

Explanation

To find the average values of kinetic energy (KE) and potential energy (PE) of a particle executing simple harmonic motion (SHM), we can follow these steps:

### Step 1: Define the equations for kinetic and potential energy

1. **Kinetic Energy (KE)**:
The kinetic energy of a particle in SHM can be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
where \( v \) is the velocity of the particle. The velocity can be derived from the displacement \( x = A \cos(\omega t + \phi) \):
\[ v = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi) \]
Therefore, substituting for \( v \):
\[ KE = \frac{1}{2} m (A \omega \sin(\omega t + \phi))^2 = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t + \phi) \]

2. **Potential Energy (PE)**:
The potential energy in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
where \( k = m \omega^2 \) (the spring constant). Substituting for \( x \):
\[ PE = \frac{1}{2} m \omega^2 (A \cos(\omega t + \phi))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t + \phi) \]

### Step 2: Calculate the average values over one period

1. **Average Kinetic Energy**:
The average kinetic energy over one period \( T \) is given by:
\[ KE_{\text{avg}} = \frac{1}{T} \int_0^T KE \, dt = \frac{1}{T} \int_0^T \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t + \phi) \, dt \]
Using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):
\[ KE_{\text{avg}} = \frac{1}{T} \cdot \frac{1}{2} m A^2 \omega^2 \int_0^T \frac{1 - \cos(2(\omega t + \phi))}{2} \, dt \]
The integral of \( \cos(2(\omega t + \phi)) \) over one period is zero, so:
\[ KE_{\text{avg}} = \frac{1}{T} \cdot \frac{1}{2} m A^2 \omega^2 \cdot \frac{T}{2} = \frac{1}{4} m A^2 \omega^2 \]

2. **Average Potential Energy**:
Similarly, for potential energy:
\[ PE_{\text{avg}} = \frac{1}{T} \int_0^T PE \, dt = \frac{1}{T} \int_0^T \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t + \phi) \, dt \]
Using the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \):
\[ PE_{\text{avg}} = \frac{1}{T} \cdot \frac{1}{2} m A^2 \omega^2 \int_0^T \frac{1 + \cos(2(\omega t + \phi))}{2} \, dt \]
Again, the integral of \( \cos(2(\omega t + \phi)) \) over one period is zero:
\[ PE_{\text{avg}} = \frac{1}{T} \cdot \frac{1}{2} m A^2 \omega^2 \cdot \frac{T}{2} = \frac{1}{4} m A^2 \omega^2 \]

### Final Result:
Thus, the average values of kinetic energy and potential energy over one period are:
\[ KE_{\text{avg}} = \frac{1}{4} m A^2 \omega^2 \]
\[ PE_{\text{avg}} = \frac{1}{4} m A^2 \omega^2 \]