The ration of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude , the distance being measured from its equilibrium position is

Options

A. `2:1`
B. `3:1`
C. `8:1`
D. `1:1`

Explanation

To find the ratio of kinetic energy (KE) to potential energy (PE) of a particle executing simple harmonic motion (SHM) at a distance equal to half its amplitude, we can follow these steps:

### Step 1: Understand the formulas for KE and PE in SHM
The kinetic energy (KE) of a particle in SHM is given by:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
where:
- \( m \) is the mass of the particle,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude,
- \( x \) is the displacement from the equilibrium position.

The potential energy (PE) is given by:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]

### Step 2: Set the displacement
In this case, we need to find the ratio when the displacement \( x \) is equal to half the amplitude, i.e., \( x = \frac{A}{2} \).

### Step 3: Substitute \( x \) into the KE and PE formulas
Substituting \( x = \frac{A}{2} \) into the KE formula:
\[ KE = \frac{1}{2} m \omega^2 \left(A^2 - \left(\frac{A}{2}\right)^2\right) \]
Calculating \( \left(\frac{A}{2}\right)^2 \):
\[ \left(\frac{A}{2}\right)^2 = \frac{A^2}{4} \]
Thus,
\[ KE = \frac{1}{2} m \omega^2 \left(A^2 - \frac{A^2}{4}\right) = \frac{1}{2} m \omega^2 \left(\frac{4A^2}{4} - \frac{A^2}{4}\right) = \frac{1}{2} m \omega^2 \left(\frac{3A^2}{4}\right) \]
So,
\[ KE = \frac{3}{8} m \omega^2 A^2 \]

Now substituting \( x = \frac{A}{2} \) into the PE formula:
\[ PE = \frac{1}{2} m \omega^2 \left(\frac{A}{2}\right)^2 = \frac{1}{2} m \omega^2 \left(\frac{A^2}{4}\right) = \frac{1}{8} m \omega^2 A^2 \]

### Step 4: Find the ratio of KE to PE
Now we can find the ratio of kinetic energy to potential energy:
\[ \frac{KE}{PE} = \frac{\frac{3}{8} m \omega^2 A^2}{\frac{1}{8} m \omega^2 A^2} \]
The \( m \), \( \omega^2 \), and \( A^2 \) terms cancel out:
\[ \frac{KE}{PE} = \frac{3}{1} = 3 \]

### Final Answer
Thus, the ratio of kinetic energy to potential energy when the displacement is half the amplitude is:
\[ \text{Ratio} = 3 : 1 \]