A particle is executing simple harmonic motion with frequency f. Match the columns.
`{:(,"Column-I",,"Column-II"),((A),"Zero",(p),"Frequency with which kinetic energy of particle osciallates"),((B),t,(q),"Frequency with which potential energy of particle oscillates"),((C ), 2f, (r ),"Frequency with which difference between kinetic and potential energy of oscillates"),((D),4f,(s),"Frequency with which velocity of paritcle osciallates "),(,,(t),"Frequency with which total mechanical energy oscillates"):}`

Answer

To solve the problem, we need to match the frequencies of various quantities related to a particle executing simple harmonic motion (SHM) with frequency \( f \). The quantities we need to analyze are:

1. Kinetic energy of the particle
2. Potential energy of the particle
3. Difference between kinetic and potential energy
4. Velocity of the particle
5. Total mechanical energy of the system

Let's analyze each of these step by step.

### Step 1: Kinetic Energy
The kinetic energy \( KE \) of a particle in SHM is given by:
\[ KE = \frac{1}{2} m v^2 \]
where \( v = A \omega \cos(\omega t) \). Thus,
\[ KE = \frac{1}{2} m (A \omega \cos(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \]
Using the identity \( \cos^2(\omega t) = \frac{1 + \cos(2\omega t)}{2} \), we can rewrite this as:
\[ KE = \frac{1}{4} m A^2 \omega^2 + \frac{1}{4} m A^2 \omega^2 \cos(2\omega t) \]
The term with \( \cos(2\omega t) \) indicates that the frequency of oscillation of the kinetic energy is \( 2\omega \), which corresponds to \( 2f \).

### Step 2: Potential Energy
The potential energy \( PE \) in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
where \( x = A \sin(\omega t) \). Thus,
\[ PE = \frac{1}{2} k (A \sin(\omega t))^2 = \frac{1}{2} k A^2 \sin^2(\omega t) \]
Using the identity \( \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} \), we can rewrite this as:
\[ PE = \frac{1}{4} k A^2 - \frac{1}{4} k A^2 \cos(2\omega t) \]
The term with \( \cos(2\omega t) \) indicates that the frequency of oscillation of the potential energy is also \( 2\omega \), which corresponds to \( 2f \).

### Step 3: Difference Between Kinetic and Potential Energy
The difference \( KE - PE \) can be expressed as:
\[ KE - PE = \left(\frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) - \frac{1}{2} k A^2 \sin^2(\omega t)\right) \]
Using \( k = m \omega^2 \), we can simplify this to:
\[ KE - PE = \frac{1}{2} A^2 \left(m \omega^2 \cos^2(\omega t) - k \sin^2(\omega t)\right) \]
This will also oscillate with frequency \( 2\omega \), corresponding to \( 2f \).

### Step 4: Velocity of the Particle
The velocity \( v \) of the particle is given by:
\[ v = A \omega \cos(\omega t) \]
The frequency of the velocity oscillation is the same as the frequency of the motion, which is \( f \).

### Step 5: Total Mechanical Energy
The total mechanical energy \( E \) in SHM is given by:
\[ E = KE + PE \]
Since both kinetic and potential energy oscillate, but their sum remains constant, the total mechanical energy does not oscillate. Therefore, the frequency of the total mechanical energy is \( 0 \).

### Final Matching
Now we can match the results with the columns given in the question:

- (A) Zero matches with (t) "Frequency with which total mechanical energy oscillates"
- (B) \( f \) matches with (s) "Frequency with which velocity of particle oscillates"
- (C) \( 2f \) matches with (p) "Frequency with which kinetic energy of particle oscillates"
- (C) \( 2f \) matches with (q) "Frequency with which potential energy of particle oscillates"
- (C) \( 2f \) matches with (r) "Frequency with which difference between kinetic and potential energy oscillates"
- (D) \( 4f \) does not match with any option.

### Summary of Matches:
- A → t
- B → s
- C → p, q, r
- D → No match