A particle of mass m performs SHM along a straight line with frequency f and amplitude A:-

Options

A. The average kinetic energy of the particle is zero
B. The average potential energy is `mpi^(2)f^(2)A^(2)`
C. The frequency of oscillation of kinetic energy is 2f
D. Velocity function leads acceleration by `pi//2`

Answer

To solve the problem step by step, we will analyze the motion of the particle performing Simple Harmonic Motion (SHM) and derive the necessary quantities.

### Step 1: Understand the SHM Equation
The displacement \( x \) of a particle in SHM can be expressed as:
\[ x(t) = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

### Step 2: Relate Angular Frequency to Frequency
The angular frequency \( \omega \) is related to the frequency \( f \) by:
\[ \omega = 2\pi f \]

### Step 3: Find the Velocity
The velocity \( v \) is the time derivative of displacement:
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(A \sin(\omega t)) = A \omega \cos(\omega t) \]
Substituting \( \omega = 2\pi f \):
\[ v(t) = A (2\pi f) \cos(2\pi f t) \]

### Step 4: Find the Acceleration
The acceleration \( a \) is the time derivative of velocity:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(A \omega \cos(\omega t)) = -A \omega^2 \sin(\omega t) \]
Substituting \( \omega = 2\pi f \):
\[ a(t) = -A (2\pi f)^2 \sin(2\pi f t) \]

### Step 5: Analyze the Phase Relationship
From the equations of velocity and acceleration:
- Velocity \( v(t) = A (2\pi f) \cos(2\pi f t) \)
- Acceleration \( a(t) = -A (2\pi f)^2 \sin(2\pi f t) \)

We see that:
- The velocity is proportional to \( \cos(\omega t) \)
- The acceleration is proportional to \( -\sin(\omega t) \)

Since \( \cos(\theta) \) leads \( -\sin(\theta) \) by \( \frac{\pi}{2} \), we conclude:
\[ \text{Velocity lags acceleration by } \frac{\pi}{2}. \]

### Step 6: Find Average Kinetic Energy
The kinetic energy \( K \) of the particle is given by:
\[ K = \frac{1}{2} mv^2 = \frac{1}{2} m (A (2\pi f) \cos(2\pi f t))^2 \]
\[ K = \frac{1}{2} m (4\pi^2 f^2 A^2 \cos^2(2\pi f t)) \]

The average value of \( \cos^2(\theta) \) over one complete cycle is \( \frac{1}{2} \):
\[ \langle K \rangle = \frac{1}{2} m (4\pi^2 f^2 A^2) \cdot \frac{1}{2} = \frac{1}{4} m (4\pi^2 f^2 A^2) = m \pi^2 f^2 A^2 \]

### Step 7: Find Average Potential Energy
The potential energy \( U \) in SHM is given by:
\[ U = \frac{1}{2} k x^2 \]
where \( k = m \omega^2 = m (2\pi f)^2 \). Thus:
\[ U = \frac{1}{2} m (2\pi f)^2 A^2 \sin^2(2\pi f t) \]

The average value of \( \sin^2(\theta) \) over one complete cycle is also \( \frac{1}{2} \):
\[ \langle U \rangle = \frac{1}{2} m (4\pi^2 f^2 A^2) \cdot \frac{1}{2} = \frac{1}{4} m (4\pi^2 f^2 A^2) = m \pi^2 f^2 A^2 \]

### Step 8: Frequency of Oscillation of Kinetic Energy
The frequency of oscillation of kinetic energy is double that of the displacement, thus:
\[ f' = 2f \]

### Final Answers
1. Velocity lags acceleration by \( \frac{\pi}{2} \).
2. Average kinetic energy: \( m \pi^2 f^2 A^2 \).
3. Average potential energy: \( m \pi^2 f^2 A^2 \).
4. Frequency of oscillation of kinetic energy: \( 2f \).