A small mass m attached to one end of a spring with a negligible mass and an unstretched length L, executes vertical oscillations with angular frequency `omega_(0)`. When the mass is rotated with an angular speed `omega` by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during this rotation is

Options

A. `(omega^(2)L)/(omega_(0)^(2)-omega^(2))`
B. `(omega_(0)^(2)L)/(omega^(2)-omega_(0)^(2))`
C. `(omega^(2)L)/(omega_(0)^(2))`
D. `(omega_(0)^(2)L)/(omega^(2))`

Answer

To solve the problem, we need to analyze the forces acting on the mass \( m \) when it is rotated in a circular path while attached to the spring. Here's a step-by-step solution:

### Step 1: Understand the system
We have a mass \( m \) attached to a spring with an unstretched length \( L \). The mass executes vertical oscillations with an angular frequency \( \omega_0 \). When the mass is rotated with an angular speed \( \omega \), it moves in a horizontal circular path.

### Step 2: Define the variables
- Let \( x \) be the extension of the spring when the mass is rotated.
- The total length of the spring when extended is \( L + x \).
- The radius \( r \) of the circular motion is given by \( r = (L + x) \sin \theta \), where \( \theta \) is the angle the spring makes with the vertical.

### Step 3: Write the forces acting on the mass
1. The spring force acting upwards is \( kx \) (where \( k \) is the spring constant).
2. The weight of the mass acting downwards is \( mg \).
3. The centrifugal force acting outward due to circular motion is \( m r \omega^2 \).

### Step 4: Set up the force balance
In the vertical direction, the forces must balance:
\[ kx \cos \theta = mg \]
In the horizontal direction, the centrifugal force is balanced by the horizontal component of the spring force:
\[ m r \omega^2 = kx \sin \theta \]

### Step 5: Substitute for \( r \)
Substituting \( r = (L + x) \sin \theta \) into the centrifugal force equation gives:
\[ m (L + x) \sin \theta \omega^2 = kx \sin \theta \]
Dividing both sides by \( \sin \theta \) (assuming \( \sin \theta \neq 0 \)):
\[ m (L + x) \omega^2 = kx \]

### Step 6: Express \( k \) in terms of \( \omega_0 \)
From the vertical oscillation, we know:
\[ \omega_0 = \sqrt{\frac{k}{m}} \implies k = m \omega_0^2 \]
Substituting this into the equation gives:
\[ m (L + x) \omega^2 = m \omega_0^2 x \]
Cancelling \( m \) (assuming \( m \neq 0 \)):
\[ (L + x) \omega^2 = \omega_0^2 x \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ L \omega^2 = x(\omega_0^2 - \omega^2) \]
Thus, we can solve for \( x \):
\[ x = \frac{L \omega^2}{\omega_0^2 - \omega^2} \]

### Conclusion
The increase in the length of the spring during rotation is:
\[ \boxed{x = \frac{L \omega^2}{\omega_0^2 - \omega^2}} \]