A particle of mass 1 kg is hanging from a spring of force constant 100 `Nm^(1)` . The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is `(T)/(x)` The value of x is_______

Answer

To solve the problem, we need to find the time at which the kinetic energy (KE) and potential energy (PE) of a mass-spring system are equal during simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Identify the parameters:**
- Mass (m) = 1 kg
- Spring constant (k) = 100 N/m

2. **Calculate the time period (T) of the SHM:**
The formula for the time period \( T \) of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting the values:
\[ T = 2\pi \sqrt{\frac{1}{100}} = 2\pi \sqrt{0.01} = 2\pi \cdot 0.1 = 0.2\pi \text{ seconds} \]

3. **Formulas for kinetic and potential energy:**
In SHM, the potential energy (PE) and kinetic energy (KE) can be expressed as:
- Potential Energy:
\[ PE = \frac{1}{2} k x^2 \]
- Kinetic Energy:
\[ KE = \frac{1}{2} m v^2 \]
where \( v \) is the velocity of the mass.

4. **Expressing PE and KE in terms of displacement:**
For SHM, the displacement \( x \) can be expressed as:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega = \frac{2\pi}{T} \).

The velocity \( v \) is given by:
\[ v = \frac{dx}{dt} = A \omega \cos(\omega t) \]

5. **Substituting into the energy equations:**
- Potential Energy becomes:
\[ PE = \frac{1}{2} k (A \sin(\omega t))^2 = \frac{1}{2} k A^2 \sin^2(\omega t) \]
- Kinetic Energy becomes:
\[ KE = \frac{1}{2} m (A \omega \cos(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \]

6. **Setting KE equal to PE:**
To find when KE = PE:
\[ \frac{1}{2} k A^2 \sin^2(\omega t) = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \]
Canceling \( \frac{1}{2} A^2 \) from both sides gives:
\[ k \sin^2(\omega t) = m \omega^2 \cos^2(\omega t) \]

7. **Using the relationship \( \omega^2 = \frac{k}{m} \):**
Substitute \( \omega^2 \):
\[ k \sin^2(\omega t) = k \sin^2(\omega t) \cos^2(\omega t) \]
This simplifies to:
\[ \sin^2(\omega t) = \cos^2(\omega t) \]
This implies:
\[ \tan^2(\omega t) = 1 \Rightarrow \tan(\omega t) = 1 \Rightarrow \omega t = \frac{\pi}{4} \]

8. **Solving for time \( t \):**
Substitute \( \omega = \frac{2\pi}{T} \):
\[ \frac{2\pi}{T} t = \frac{\pi}{4} \Rightarrow t = \frac{T}{8} \]

9. **Finding the value of \( x \):**
Since \( t = \frac{T}{x} \), we have:
\[ x = 8 \]

### Final Answer:
The value of \( x \) is **8**.