In a S.H.M. with amplitude 'A', what is the ratio of K.E. and P.E. at `(A)/(2)` distance from the mean position ?

Options

A. `1:4`
B. `3:4`
C. `3:1`
D. `1:3`

Answer

To find the ratio of kinetic energy (K.E.) and potential energy (P.E.) at a distance of \( \frac{A}{2} \) from the mean position in simple harmonic motion (S.H.M.), we can follow these steps:

### Step 1: Understand the formulas for K.E. and P.E.
In S.H.M., the kinetic energy (K.E.) and potential energy (P.E.) can be expressed as:
- K.E. = \( \frac{1}{2} m \omega^2 (A^2 - x^2) \)
- P.E. = \( \frac{1}{2} m \omega^2 x^2 \)

Where:
- \( m \) = mass of the oscillating object
- \( \omega \) = angular frequency
- \( A \) = amplitude
- \( x \) = displacement from the mean position

### Step 2: Substitute the value of \( x \)
We need to find the K.E. and P.E. when \( x = \frac{A}{2} \).

### Step 3: Calculate K.E. at \( x = \frac{A}{2} \)
Substituting \( x = \frac{A}{2} \) into the K.E. formula:
\[ K.E. = \frac{1}{2} m \omega^2 \left( A^2 - \left( \frac{A}{2} \right)^2 \right) \]
\[ = \frac{1}{2} m \omega^2 \left( A^2 - \frac{A^2}{4} \right) \]
\[ = \frac{1}{2} m \omega^2 \left( \frac{4A^2}{4} - \frac{A^2}{4} \right) \]
\[ = \frac{1}{2} m \omega^2 \left( \frac{3A^2}{4} \right) \]
\[ = \frac{3}{8} m \omega^2 A^2 \]

### Step 4: Calculate P.E. at \( x = \frac{A}{2} \)
Substituting \( x = \frac{A}{2} \) into the P.E. formula:
\[ P.E. = \frac{1}{2} m \omega^2 \left( \frac{A}{2} \right)^2 \]
\[ = \frac{1}{2} m \omega^2 \left( \frac{A^2}{4} \right) \]
\[ = \frac{1}{8} m \omega^2 A^2 \]

### Step 5: Find the ratio of K.E. to P.E.
Now we find the ratio of K.E. to P.E.:
\[ \text{Ratio} = \frac{K.E.}{P.E.} = \frac{\frac{3}{8} m \omega^2 A^2}{\frac{1}{8} m \omega^2 A^2} \]
The \( m \), \( \omega^2 \), and \( A^2 \) terms cancel out:
\[ = \frac{3}{1} = 3 \]

### Final Answer
The ratio of kinetic energy to potential energy at a distance of \( \frac{A}{2} \) from the mean position is \( 3:1 \).

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