A body is executing `SHM` under the action of force whose maximum magnitude is `5sqrt(2)N`. The magnitude of force acting on the particle at the instant when its kintentic energy and potential are euqal is `10x`. Find `x`. (Assume potenital energy to be zero at mean position)

Explanation

To solve the problem step by step, we need to analyze the given information about a body executing Simple Harmonic Motion (SHM).

### Step 1: Understand the Maximum Force in SHM
The maximum force \( F_{\text{max}} \) in SHM is given by:
\[ F_{\text{max}} = m \omega^2 A \]
where \( A \) is the amplitude of the motion. We are given that \( F_{\text{max}} = 5\sqrt{2} \, \text{N} \).

### Step 2: Relate Kinetic Energy and Potential Energy
In SHM, when the kinetic energy (KE) is equal to the potential energy (PE), we can express them as:
\[ KE = \frac{1}{2} m v^2 \]
\[ PE = \frac{1}{2} k x^2 \]
where \( k = m \omega^2 \) and \( x \) is the displacement from the mean position.

Since \( KE = PE \), we can set the equations equal:
\[ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \]
This simplifies to:
\[ m v^2 = k x^2 \]

### Step 3: Express Velocity in Terms of Amplitude
In SHM, the velocity \( v \) can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
Substituting this into the kinetic energy equation gives:
\[ m (\omega \sqrt{A^2 - x^2})^2 = k x^2 \]
Substituting \( k = m \omega^2 \):
\[ m \omega^2 (A^2 - x^2) = m \omega^2 x^2 \]
Cancelling \( m \omega^2 \) (assuming \( m \neq 0 \) and \( \omega \neq 0 \)):
\[ A^2 - x^2 = x^2 \]
This simplifies to:
\[ A^2 = 2x^2 \]
Thus, we can express \( x \) in terms of \( A \):
\[ x = \frac{A}{\sqrt{2}} \]

### Step 4: Calculate the Force at \( x \)
The force \( F \) at displacement \( x \) is given by:
\[ F = -k x = -m \omega^2 x \]
Substituting \( x = \frac{A}{\sqrt{2}} \):
\[ F = -m \omega^2 \left(\frac{A}{\sqrt{2}}\right) = -\frac{m \omega^2 A}{\sqrt{2}} \]

### Step 5: Relate to Given Force
We know that the maximum force \( F_{\text{max}} = m \omega^2 A = 5\sqrt{2} \, \text{N} \). Therefore:
\[ F = -\frac{5\sqrt{2}}{\sqrt{2}} = -5 \, \text{N} \]

### Step 6: Set Up the Equation
According to the problem, the magnitude of the force at this instant is given as \( 10x \):
\[ 10x = 5 \]

### Step 7: Solve for \( x \)
From the equation \( 10x = 5 \):
\[ x = \frac{5}{10} = 0.5 \]

Thus, the value of \( x \) is:
\[ \boxed{0.5} \]