A particle of mass `m` executes `SHM` with amplitude 'a' and frequency 'v'. The average kinetic energy during motion from the position of equilibrium to the end is:

Options

A. (a)`1/4m a^(2)v^(2)`
B. (b)`4pi^(2)m a^(2)v^(2)`
C. (c)`2pi^(2)m a^(2)v^(2)`
D. (d)`pi^(2)m a^(2)v^(2)`

Explanation

To find the average kinetic energy of a particle executing Simple Harmonic Motion (SHM) from the position of equilibrium to the end, we can follow these steps:

### Step 1: Understand the Kinetic Energy in SHM
The kinetic energy (KE) of a particle in SHM is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

### Step 2: Determine the Velocity in SHM
In SHM, the displacement \( y \) can be expressed as:
\[ y = a \sin(\omega t) \]
where \( a \) is the amplitude and \( \omega \) is the angular frequency. The velocity \( v \) can be derived by differentiating the displacement with respect to time:
\[ v = \frac{dy}{dt} = a \omega \cos(\omega t) \]

### Step 3: Substitute Velocity into the Kinetic Energy Formula
Substituting the expression for velocity into the kinetic energy formula:
\[ KE = \frac{1}{2} m (a \omega \cos(\omega t))^2 \]
\[ KE = \frac{1}{2} m a^2 \omega^2 \cos^2(\omega t) \]

### Step 4: Calculate Average Kinetic Energy
To find the average kinetic energy from the position of equilibrium (where \( y = 0 \)) to the maximum displacement (where \( y = a \)), we need to integrate the kinetic energy over this interval and then divide by the time period of the motion.

The average kinetic energy \( \langle KE \rangle \) can be calculated as:
\[ \langle KE \rangle = \frac{1}{T} \int_0^T KE \, dt \]
However, we can simplify this by noting that the average value of \( \cos^2(\omega t) \) over one complete cycle is \( \frac{1}{2} \).

Thus, the average kinetic energy from equilibrium to maximum displacement is:
\[ \langle KE \rangle = \frac{1}{2} m a^2 \omega^2 \cdot \frac{1}{2} = \frac{1}{4} m a^2 \omega^2 \]

### Step 5: Relate Angular Frequency to Frequency
The angular frequency \( \omega \) is related to the frequency \( v \) by:
\[ \omega = 2\pi v \]
Substituting this into the average kinetic energy expression gives:
\[ \langle KE \rangle = \frac{1}{4} m a^2 (2\pi v)^2 \]
\[ \langle KE \rangle = \frac{1}{4} m a^2 (4\pi^2 v^2) \]
\[ \langle KE \rangle = m a^2 \pi^2 v^2 \]

### Final Result
Thus, the average kinetic energy during motion from the position of equilibrium to the end is:
\[ \langle KE \rangle = \frac{1}{4} m a^2 \omega^2 \]