A body is executing SHM with amplitude a and time period T. The ratio of kinetic and potential energy when displacement from the equilibrium position is half the amplitude

Options

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

Answer

To solve the problem of finding the ratio of kinetic energy (KE) to potential energy (PE) when the displacement from the equilibrium position is half the amplitude in simple harmonic motion (SHM), we can follow these steps:

### Step 1: Understand the parameters
- Let the amplitude be \( A \).
- The displacement from the equilibrium position is given as \( x = \frac{A}{2} \).
- The time period is \( T \), but it is not directly needed for this calculation.

### Step 2: Write the formulas for kinetic and potential energy
1. **Kinetic Energy (KE)** in SHM is given by:
\[ KE = \frac{1}{2} k (A^2 - x^2) \]
where \( k \) is the spring constant.

2. **Potential Energy (PE)** in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]

### Step 3: Substitute \( x = \frac{A}{2} \) into the formulas
1. **Calculate KE**:
\[ KE = \frac{1}{2} k \left( A^2 - \left( \frac{A}{2} \right)^2 \right) \]
\[ = \frac{1}{2} k \left( A^2 - \frac{A^2}{4} \right) \]
\[ = \frac{1}{2} k \left( \frac{4A^2}{4} - \frac{A^2}{4} \right) \]
\[ = \frac{1}{2} k \left( \frac{3A^2}{4} \right) \]
\[ = \frac{3}{8} k A^2 \]

2. **Calculate PE**:
\[ PE = \frac{1}{2} k \left( \frac{A}{2} \right)^2 \]
\[ = \frac{1}{2} k \left( \frac{A^2}{4} \right) \]
\[ = \frac{1}{8} k A^2 \]

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

### Final Answer
The ratio of kinetic energy to potential energy when the displacement from the equilibrium position is half the amplitude is:
\[ \text{Ratio of KE to PE} = 3:1 \]