A piece of wire is bent in the shape of a parabola `y=kx^(2)` (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is:

Options

A. (a)`(a)/(gk)`
B. (b)`(a)/(2 gk)`
C. (c)`(2a)/(gk)`
D. (d)`(a)/(4 gk)`

Explanation

To solve the problem, we need to analyze the forces acting on the bead when the wire is accelerated. We will follow these steps:

### Step 1: Understand the System
We have a parabola defined by the equation \( y = kx^2 \). The bead of mass \( m \) can slide on this wire without friction. Initially, when the wire is at rest, the bead is at the lowest point of the parabola, which is at the origin (0,0).

### Step 2: Analyze the Forces
When the wire is accelerated to the right with a constant acceleration \( a \), the bead will experience two forces:
1. The gravitational force acting downward: \( F_g = mg \)
2. The normal force \( N \) exerted by the wire on the bead.

### Step 3: Determine the Angle of the Wire
The slope of the parabola at any point \( x \) is given by the derivative of \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2kx \]
The angle \( \theta \) that the tangent to the parabola makes with the horizontal can be expressed using the tangent function:
\[ \tan \theta = \frac{dy}{dx} = 2kx \]

### Step 4: Resolve Forces
The normal force \( N \) can be resolved into two components:
- The vertical component: \( N \cos \theta \)
- The horizontal component: \( N \sin \theta \)

From the equilibrium of forces in the vertical direction, we have:
\[ N \cos \theta = mg \]

In the horizontal direction, since the wire is accelerating, we have:
\[ N \sin \theta = ma \]

### Step 5: Relate the Forces
Using the expressions for \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{2kx}{\sqrt{1 + (2kx)^2}}, \quad \cos \theta = \frac{1}{\sqrt{1 + (2kx)^2}} \]

Substituting these into the force equations gives:
1. \( N \frac{1}{\sqrt{1 + (2kx)^2}} = mg \)
2. \( N \frac{2kx}{\sqrt{1 + (2kx)^2}} = ma \)

### Step 6: Eliminate \( N \)
From the first equation, we can express \( N \):
\[ N = mg \sqrt{1 + (2kx)^2} \]

Substituting \( N \) into the second equation:
\[ mg \sqrt{1 + (2kx)^2} \cdot \frac{2kx}{\sqrt{1 + (2kx)^2}} = ma \]
This simplifies to:
\[ mg \cdot 2kx = ma \]

### Step 7: Solve for \( x \)
Rearranging gives:
\[ x = \frac{a}{2g} \cdot \frac{1}{k} \]

### Final Result
The distance of the new equilibrium position of the bead from the y-axis is:
\[ x = \frac{a}{2gk} \]