The ratio of the K.E. and P.E. possessed by a body executing S.H.M. when it is at a distance of `1/n` of its amplitude from the mean position, is

Options

A. `n^2`
B. `1/2n^2`
C. `n^2+1`
D. `n^2-1`

Answer

To find the ratio of kinetic energy (K.E.) and potential energy (P.E.) for a body executing simple harmonic motion (S.H.M.) when it is at a distance of \( \frac{1}{n} \) of its amplitude from the mean position, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( A \) = Amplitude of the motion
- \( x \) = Displacement from the mean position
- \( m \) = Mass of the body
- \( \omega \) = Angular frequency

Given that the body is at a distance of \( \frac{1}{n} \) of its amplitude from the mean position:
\[ x = \frac{A}{n} \]

### Step 2: Write the Expressions for K.E. and P.E.
The kinetic energy (K.E.) and potential energy (P.E.) in S.H.M. are given by the following formulas:

1. K.E.:
\[ \text{K.E.} = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

2. P.E.:
\[ \text{P.E.} = \frac{1}{2} m \omega^2 x^2 \]

### Step 3: Substitute the Value of \( x \)
Substituting \( x = \frac{A}{n} \) into the expressions for K.E. and P.E.:

- For K.E.:
\[ \text{K.E.} = \frac{1}{2} m \omega^2 \left( A^2 - \left( \frac{A}{n} \right)^2 \right) = \frac{1}{2} m \omega^2 \left( A^2 - \frac{A^2}{n^2} \right) \]
\[ = \frac{1}{2} m \omega^2 A^2 \left( 1 - \frac{1}{n^2} \right) \]

- For P.E.:
\[ \text{P.E.} = \frac{1}{2} m \omega^2 \left( \frac{A}{n} \right)^2 = \frac{1}{2} m \omega^2 \frac{A^2}{n^2} \]

### Step 4: Find the Ratio of K.E. to P.E.
Now we can find the ratio of K.E. to P.E.:
\[ \frac{\text{K.E.}}{\text{P.E.}} = \frac{\frac{1}{2} m \omega^2 A^2 \left( 1 - \frac{1}{n^2} \right)}{\frac{1}{2} m \omega^2 \frac{A^2}{n^2}} \]

The \( \frac{1}{2} m \omega^2 A^2 \) terms cancel out:
\[ = \frac{1 - \frac{1}{n^2}}{\frac{1}{n^2}} = \left( 1 - \frac{1}{n^2} \right) \cdot n^2 \]

### Step 5: Simplify the Expression
Now, simplify the expression:
\[ = n^2 - 1 \]

### Final Answer
Thus, the ratio of kinetic energy to potential energy when the body is at a distance of \( \frac{1}{n} \) of its amplitude from the mean position is:
\[ \frac{\text{K.E.}}{\text{P.E.}} = n^2 - 1 \]