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 find 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 of SHM
- The amplitude of the motion is given as \( A \).
- The displacement from the equilibrium position is given as \( x = \frac{A}{2} \) (half the amplitude).
- The time period is \( T \), but it is not directly needed for this calculation.

### Step 2: Write the expression for kinetic energy
The kinetic energy (KE) of a body in SHM is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( v \) is the velocity of the body. The velocity \( v \) can be expressed in terms of angular frequency \( \omega \) and displacement \( x \):
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( \omega = \frac{2\pi}{T} \).

### Step 3: Substitute for velocity at \( x = \frac{A}{2} \)
Substituting \( x = \frac{A}{2} \) into the velocity formula:
\[ v = \omega \sqrt{A^2 - \left(\frac{A}{2}\right)^2} = \omega \sqrt{A^2 - \frac{A^2}{4}} = \omega \sqrt{\frac{3A^2}{4}} = \frac{\omega A \sqrt{3}}{2} \]
Now, substituting this into the kinetic energy formula:
\[ KE = \frac{1}{2} m \left(\frac{\omega A \sqrt{3}}{2}\right)^2 = \frac{1}{2} m \cdot \frac{3 \omega^2 A^2}{4} = \frac{3}{8} m \omega^2 A^2 \]

### Step 4: Write the expression for potential energy
The potential energy (PE) in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
Using \( \omega = \sqrt{\frac{k}{m}} \), we can express \( k \) as \( k = m \omega^2 \). Thus,
\[ PE = \frac{1}{2} m \omega^2 x^2 \]
Substituting \( x = \frac{A}{2} \):
\[ PE = \frac{1}{2} m \omega^2 \left(\frac{A}{2}\right)^2 = \frac{1}{2} m \omega^2 \cdot \frac{A^2}{4} = \frac{1}{8} m \omega^2 A^2 \]

### 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:
\[ \text{Ratio} = \frac{3}{1} = 3 \]

### Final Answer
The ratio of kinetic energy to potential energy when the displacement is half the amplitude is \( 3:1 \).

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