For a two body oscillator system, prove the relation,
`T = 2pi sqrt((mu)/(k))`
where, `mu = (m_(1)m_(2))/(m_(1) + (m_(2))) =` reduced mass.

Explanation

To prove the relation \( T = 2\pi \sqrt{\frac{\mu}{k}} \) for a two-body oscillator system, where \( \mu = \frac{m_1 m_2}{m_1 + m_2} \) is the reduced mass, we will follow these steps:

### Step 1: Understand the System
We have two masses \( m_1 \) and \( m_2 \) connected by a spring with spring constant \( k \). When the system is displaced, both masses will experience forces due to the spring.

### Step 2: Define the Displacement
Let \( x_1 \) be the displacement of mass \( m_1 \) and \( x_2 \) be the displacement of mass \( m_2 \). The total extension of the spring can be expressed as:
\[ x = x_1 - x_2 - L_0 \]
where \( L_0 \) is the natural length of the spring.

### Step 3: Apply Newton's Second Law
According to Newton's second law, the force on each mass due to the spring can be expressed as:
\[ F = -kx \]
For mass \( m_1 \):
\[ m_1 \frac{d^2 x_1}{dt^2} = -k(x_1 - x_2 - L_0) \]
For mass \( m_2 \):
\[ m_2 \frac{d^2 x_2}{dt^2} = -k(x_2 - x_1 + L_0) \]

### Step 4: Combine the Equations
Rearranging the equations gives us two second-order differential equations. We can express the relative displacement \( x = x_1 - x_2 \).

### Step 5: Formulate the Differential Equation
By substituting \( x_1 \) and \( x_2 \) into the equations and eliminating one of the variables, we can derive a single equation in terms of \( x \):
\[ \mu \frac{d^2 x}{dt^2} = -kx \]
where \( \mu = \frac{m_1 m_2}{m_1 + m_2} \) is the reduced mass.

### Step 6: Identify the Form of the Equation
The equation \( \mu \frac{d^2 x}{dt^2} = -kx \) is a standard form of the simple harmonic motion equation, which can be written as:
\[ \frac{d^2 x}{dt^2} + \frac{k}{\mu} x = 0 \]

### Step 7: Determine the Time Period
The general solution for the time period \( T \) of simple harmonic motion is given by:
\[ T = 2\pi \sqrt{\frac{\text{mass}}{\text{spring constant}}} \]
In our case, substituting \( \mu \) for mass and \( k \) for spring constant gives:
\[ T = 2\pi \sqrt{\frac{\mu}{k}} \]

### Conclusion
Thus, we have proved that the time period \( T \) of a two-body oscillator system is given by:
\[ T = 2\pi \sqrt{\frac{\mu}{k}} \]

---