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A piece of wire is bent in the shape of a parabola y= kx² (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 stays at rest with respect to the wire, from the y-axis is
A) a/gk
B) a/2gk
C) 2a/gk
D) a/4gk?
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A piece of wire is bent in the shape of a parabola y= kx² (y-axis vert...


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A piece of wire is bent in the shape of a parabola y= kx² (y-axis vert...
Understanding the Problem
When the wire is accelerated parallel to the x-axis with a constant acceleration 'a', the bead on the wire experiences a change in equilibrium due to the inertial effects.
Forces Acting on the Bead
- The bead is subject to gravitational force acting downwards (mg).
- In the accelerating frame, a pseudo-force acts on the bead in the opposite direction of the acceleration (−ma) due to the wire's acceleration.
Equation of Motion
- The bead will find a new equilibrium position where the net force acting on it is zero.
- The forces can be balanced in the vertical and horizontal directions.
Finding the New Equilibrium Position
- The effective force acting on the bead can be represented as:
- In the x-direction: ma (pseudo-force due to acceleration) = mg * (slope of the parabola).
- The slope of the parabola at any point (x, y) is given by dy/dx = 2kx, where y = kx².
Setting Up the Equation
- At equilibrium, ma = mg * (2kx).
- Rearranging gives: x = a/(2gk).
Conclusion
- Thus, the distance of the new equilibrium position of the bead from the y-axis is:
Final Answer: B) a/2gk
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A piece of wire is bent in the shape of a parabola y= kx² (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 stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk?
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A piece of wire is bent in the shape of a parabola y= kx² (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 stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A piece of wire is bent in the shape of a parabola y= kx² (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 stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A piece of wire is bent in the shape of a parabola y= kx² (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 stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk?.
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