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) for a particle of mass \( m \) executing simple harmonic motion (SHM) with amplitude \( A \) and angular frequency \( \omega \), we can follow these steps:

### Step 1: Write the expressions for kinetic and potential energy
The displacement \( x \) of the particle in SHM can be expressed as:
\[ x(t) = A \cos(\omega t + \phi) \]
The velocity \( v \) is the time derivative of displacement:
\[ v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi) \]
The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} m (A \omega \sin(\omega t + \phi))^2 = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t + \phi) \]
The potential energy \( PE \) is given by:
\[ PE = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 (A \cos(\omega t + \phi))^2 = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t + \phi) \]

### Step 2: Calculate the average kinetic energy over one period
The average kinetic energy \( \langle KE \rangle \) over one complete cycle \( T \) is given by:
\[ \langle KE \rangle = \frac{1}{T} \int_0^T KE \, dt \]
Substituting the expression for \( KE \):
\[ \langle KE \rangle = \frac{1}{T} \int_0^T \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t + \phi) \, dt \]
Using the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \):
\[ \langle KE \rangle = \frac{1}{T} \int_0^T \frac{1}{2} m A^2 \omega^2 \cdot \frac{1 - \cos(2(\omega t + \phi))}{2} \, dt \]
This simplifies to:
\[ \langle KE \rangle = \frac{m A^2 \omega^2}{4T} \int_0^T (1 - \cos(2(\omega t + \phi))) \, dt \]
The integral of \( \cos \) over one complete cycle is zero, so:
\[ \langle KE \rangle = \frac{m A^2 \omega^2}{4T} \cdot T = \frac{m A^2 \omega^2}{4} \]

### Step 3: Calculate the average potential energy over one period
Similarly, the average potential energy \( \langle PE \rangle \) is given by:
\[ \langle PE \rangle = \frac{1}{T} \int_0^T PE \, dt \]
Substituting the expression for \( PE \):
\[ \langle PE \rangle = \frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t + \phi) \, dt \]
Using the identity \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \):
\[ \langle PE \rangle = \frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \cdot \frac{1 + \cos(2(\omega t + \phi))}{2} \, dt \]
This simplifies to:
\[ \langle PE \rangle = \frac{m A^2 \omega^2}{4T} \int_0^T (1 + \cos(2(\omega t + \phi))) \, dt \]
Again, the integral of \( \cos \) over one complete cycle is zero, so:
\[ \langle PE \rangle = \frac{m A^2 \omega^2}{4T} \cdot T = \frac{m A^2 \omega^2}{4} \]

### Final Result
Thus, the average values of kinetic energy and potential energy over a period are:
\[ \langle KE \rangle = \frac{m A^2 \omega^2}{4} \]
\[ \langle PE \rangle = \frac{m A^2 \omega^2}{4} \]