When the kinetic energy of a body executing SHM is 1/3rd of the potential energy. The displacement of the body is x percent of amplitude, where x is

Options

A. `51%`
B. `87%`
C. `73%`
D. `13%`

Explanation

To solve the problem where the kinetic energy (KE) of a body executing simple harmonic motion (SHM) is one third of its potential energy (PE), we can follow these steps:

### Step 1: Write the expressions for kinetic energy and potential energy in SHM.
The kinetic energy (KE) of a body in SHM is given by:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
where \( m \) is the mass, \( \omega \) is the angular frequency, \( A \) is the amplitude, and \( x \) is the displacement.

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

### Step 2: Set up the equation based on the given condition.
According to the problem, the kinetic energy is one third of the potential energy:
\[ KE = \frac{1}{3} PE \]

Substituting the expressions for KE and PE into this equation gives:
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{3} \left( \frac{1}{2} m \omega^2 x^2 \right) \]

### Step 3: Simplify the equation.
We can cancel \(\frac{1}{2} m \omega^2\) from both sides (assuming \( m \) and \( \omega \) are not zero):
\[ A^2 - x^2 = \frac{1}{3} x^2 \]

### Step 4: Rearrange the equation.
Rearranging gives:
\[ A^2 = x^2 + \frac{1}{3} x^2 \]
\[ A^2 = \frac{4}{3} x^2 \]

### Step 5: Solve for \(x^2\).
From the equation \(A^2 = \frac{4}{3} x^2\), we can express \(x^2\) as:
\[ x^2 = \frac{3}{4} A^2 \]

### Step 6: Find \(x\).
Taking the square root of both sides gives:
\[ x = A \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} A \]

### Step 7: Calculate the percentage of amplitude.
To find \(x\) as a percentage of the amplitude \(A\):
\[ \frac{x}{A} \times 100 = \frac{\frac{\sqrt{3}}{2} A}{A} \times 100 = \frac{\sqrt{3}}{2} \times 100 \]

### Step 8: Substitute the value of \(\sqrt{3}\).
Using the approximate value \(\sqrt{3} \approx 1.732\):
\[ \frac{\sqrt{3}}{2} \times 100 \approx \frac{1.732}{2} \times 100 \approx 86.6\% \]

### Step 9: Round to the nearest option.
Rounding \(86.6\%\) gives approximately \(87\%\).

### Conclusion:
Thus, the displacement of the body is approximately \(87\%\) of the amplitude.