Starting from the origin a body oscillates simple harmonically with a period of 2s . After time `(1)/(x)` second willthe kinetic energy be `75%` of its total energy , then value of x is

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between kinetic energy and total energy in simple harmonic motion (SHM).
In SHM, the total mechanical energy (E) is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

The kinetic energy (K.E) at a displacement \( x \) is given by:
\[ K.E = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

### Step 2: Set up the equation for kinetic energy being 75% of total energy.
We know that at time \( \frac{1}{x} \) seconds, the kinetic energy is 75% of the total energy. Therefore, we can write:
\[ K.E = 0.75 E \]
Substituting the expressions for kinetic energy and total energy, we get:
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = 0.75 \left(\frac{1}{2} m \omega^2 A^2\right) \]

### Step 3: Simplify the equation.
Cancelling \( \frac{1}{2} m \omega^2 \) from both sides, we have:
\[ A^2 - x^2 = 0.75 A^2 \]

### Step 4: Rearranging the equation.
Rearranging gives:
\[ A^2 - 0.75 A^2 = x^2 \]
\[ 0.25 A^2 = x^2 \]

### Step 5: Solve for \( x \).
Taking the square root of both sides, we find:
\[ x = \sqrt{0.25} A = \frac{A}{2} \]

### Step 6: Find the value of \( x \).
Since the question asks for \( x \) in terms of the time period, we need to relate it back to the time period. The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
where \( T = 2 \) seconds. Thus,
\[ \omega = \frac{2\pi}{2} = \pi \text{ rad/s} \]

The displacement \( x \) after a time \( t \) is given by:
\[ x = A \sin(\omega t) \]

Given that \( t = \frac{1}{x} \), we can substitute this into our equation for \( x \):
\[ x = A \sin\left(\pi \cdot \frac{1}{x}\right) \]

### Conclusion
From the previous steps, we found that:
\[ x = \frac{A}{2} \]
Thus, the value of \( x \) is \( 2 \).