Ratio of kinetic energy at mean position to potential energy at A/2 of a particle performing SHM

Options

A. ` 2:1`
B. `4:1`
C. `8:1`
D. `1:1`

Explanation

To solve the problem of finding the ratio of kinetic energy at the mean position to potential energy at \( A/2 \) of a particle performing Simple Harmonic Motion (SHM), we will follow these steps:

### Step 1: Understand the Kinetic Energy at the Mean Position
In SHM, the kinetic energy (KE) at the mean position (where displacement \( y = 0 \)) is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
At the mean position, the velocity \( v \) is maximum and can be expressed as:
\[ v = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency. Thus, the kinetic energy at the mean position becomes:
\[ KE = \frac{1}{2} m (A \omega)^2 = \frac{1}{2} m A^2 \omega^2 \]

### Step 2: Calculate the Potential Energy at \( y = A/2 \)
The potential energy (PE) in SHM at a displacement \( y \) is given by:
\[ PE = \frac{1}{2} k y^2 \]
where \( k \) is the spring constant. We can relate \( k \) to \( \omega \) using:
\[ \omega = \sqrt{\frac{k}{m}} \implies k = m \omega^2 \]
Substituting this into the potential energy formula gives:
\[ PE = \frac{1}{2} (m \omega^2) \left(\frac{A}{2}\right)^2 = \frac{1}{2} m \omega^2 \frac{A^2}{4} = \frac{1}{8} m A^2 \omega^2 \]

### Step 3: Find the Ratio of Kinetic Energy to Potential Energy
Now, we can find the ratio of the kinetic energy at the mean position to the potential energy at \( y = A/2 \):
\[ \text{Ratio} = \frac{KE}{PE} = \frac{\frac{1}{2} m A^2 \omega^2}{\frac{1}{8} m A^2 \omega^2} \]
The \( m A^2 \omega^2 \) terms cancel out, leading to:
\[ \text{Ratio} = \frac{\frac{1}{2}}{\frac{1}{8}} = \frac{1}{2} \times \frac{8}{1} = 4 \]

### Conclusion
Thus, the ratio of kinetic energy at the mean position to potential energy at \( A/2 \) is:
\[ \text{Ratio} = 4 \]