When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is

Options

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

Explanation

To find the ratio of kinetic energy (KE) to potential energy (PE) of a particle executing simple harmonic motion (SHM) when its displacement is half of 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.

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

### Step 2: Substitute \(x = \frac{A}{2}\)
Since we need to find the ratio when the displacement \(x\) is half of the amplitude, we set:
\[ x = \frac{A}{2} \]

### Step 3: Calculate KE at \(x = \frac{A}{2}\)
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,
\[ KE = \frac{1}{2} m \omega^2 \left(A^2 - \frac{A^2}{4}\right) = \frac{1}{2} m \omega^2 \left(\frac{4A^2}{4} - \frac{A^2}{4}\right) = \frac{1}{2} m \omega^2 \left(\frac{3A^2}{4}\right) \]
This simplifies to:
\[ KE = \frac{3}{8} m \omega^2 A^2 \]

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

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