Show that the angle made by the string with the vertical in a conical pendulum is given by `costheta=(g)/(Lomega^(2))` , where `L` is the string and `omega` is the angular speed.

Explanation

To show that the angle made by the string with the vertical in a conical pendulum is given by the equation:

\[ \cos \theta = \frac{g}{L \omega^2} \]

where \( L \) is the length of the string and \( \omega \) is the angular speed, we can follow these steps:

### Step 1: Understanding the Setup
Consider a conical pendulum where a mass \( m \) is attached to a string of length \( L \). The mass moves in a horizontal circle while the string makes an angle \( \theta \) with the vertical.

### Step 2: Forces Acting on the Mass
The forces acting on the mass are:
1. The gravitational force \( \vec{F_g} = mg \) acting downwards.
2. The tension \( \vec{T} \) in the string acting along the string at an angle \( \theta \) to the vertical.

### Step 3: Resolving Forces
We can resolve the tension \( T \) into two components:
- Vertical component: \( T \cos \theta \)
- Horizontal component: \( T \sin \theta \)

### Step 4: Applying Newton's Second Law
In the vertical direction, the forces must balance since there is no vertical acceleration:
\[ T \cos \theta = mg \quad \text{(1)} \]

In the horizontal direction, the mass is undergoing circular motion, so we apply the centripetal force equation:
\[ T \sin \theta = \frac{mv^2}{r} \quad \text{(2)} \]
where \( r \) is the radius of the circular path.

### Step 5: Relating Radius to Angle
The radius \( r \) can be expressed in terms of the length of the string and the angle \( \theta \):
\[ r = L \sin \theta \]

### Step 6: Substituting for \( r \)
Substituting \( r \) into equation (2):
\[ T \sin \theta = \frac{mv^2}{L \sin \theta} \]

### Step 7: Expressing Velocity in Terms of Angular Speed
The linear velocity \( v \) can be related to the angular speed \( \omega \):
\[ v = r \omega = L \sin \theta \cdot \omega \]
Substituting this into the equation gives:
\[ T \sin \theta = \frac{m(L \sin \theta \cdot \omega)^2}{L \sin \theta} \]
This simplifies to:
\[ T \sin \theta = m L \sin \theta \cdot \omega^2 \]

### Step 8: Cancelling \( \sin \theta \)
Assuming \( \sin \theta \neq 0 \), we can cancel \( \sin \theta \) from both sides:
\[ T = m L \omega^2 \quad \text{(3)} \]

### Step 9: Substituting \( T \) in Equation (1)
Now substitute equation (3) into equation (1):
\[ m L \omega^2 \cos \theta = mg \]

### Step 10: Simplifying the Equation
Dividing both sides by \( m \):
\[ L \omega^2 \cos \theta = g \]

### Step 11: Final Rearrangement
Rearranging gives:
\[ \cos \theta = \frac{g}{L \omega^2} \]

This is the required proof.