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A particle free to move along the x-axis has potential energy given by U(x) = k [1–exp(–x2)] for -∞ < x < + ∞, where k is a positive constant of appropriate dimensions. Thena)at points away from the origin, the particle is in unstable equilibriumb)for any finite nonzero value of x, there is a force directed away from the originc)if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)for small displacements from x = 0, the motion is simple harmonicCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle free to move along the x-axis has potential energy given by U(x) = k [1–exp(–x2)] for -∞ < x < + ∞, where k is a positive constant of appropriate dimensions. Thena)at points away from the origin, the particle is in unstable equilibriumb)for any finite nonzero value of x, there is a force directed away from the originc)if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)for small displacements from x = 0, the motion is simple harmonicCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle free to move along the x-axis has potential energy given by U(x) = k [1–exp(–x2)] for -∞ < x < + ∞, where k is a positive constant of appropriate dimensions. Thena)at points away from the origin, the particle is in unstable equilibriumb)for any finite nonzero value of x, there is a force directed away from the originc)if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)for small displacements from x = 0, the motion is simple harmonicCorrect answer is option 'D'. Can you explain this answer?.

Solutions for A particle free to move along the x-axis has potential energy given by U(x) = k [1–exp(–x2)] for -∞ < x < + ∞, where k is a positive constant of appropriate dimensions. Thena)at points away from the origin, the particle is in unstable equilibriumb)for any finite nonzero value of x, there is a force directed away from the originc)if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)for small displacements from x = 0, the motion is simple harmonicCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of A particle free to move along the x-axis has potential energy given by U(x) = k [1–exp(–x2)] for -∞ < x < + ∞, where k is a positive constant of appropriate dimensions. Thena)at points away from the origin, the particle is in unstable equilibriumb)for any finite nonzero value of x, there is a force directed away from the originc)if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)for small displacements from x = 0, the motion is simple harmonicCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A particle free to move along the x-axis has potential energy given by U(x) = k [1–exp(–x2)] for -∞ < x < + ∞, where k is a positive constant of appropriate dimensions. Thena)at points away from the origin, the particle is in unstable equilibriumb)for any finite nonzero value of x, there is a force directed away from the originc)if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)for small displacements from x = 0, the motion is simple harmonicCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A particle free to move along the x-axis has potential energy given by U(x) = k [1–exp(–x2)] for -∞ < x < + ∞, where k is a positive constant of appropriate dimensions. Thena)at points away from the origin, the particle is in unstable equilibriumb)for any finite nonzero value of x, there is a force directed away from the originc)if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)for small displacements from x = 0, the motion is simple harmonicCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A particle free to move along the x-axis has potential energy given by U(x) = k [1–exp(–x2)] for -∞ < x < + ∞, where k is a positive constant of appropriate dimensions. Thena)at points away from the origin, the particle is in unstable equilibriumb)for any finite nonzero value of x, there is a force directed away from the originc)if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)for small displacements from x = 0, the motion is simple harmonicCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A particle free to move along the x-axis has potential energy given by U(x) = k [1–exp(–x2)] for -∞ < x < + ∞, where k is a positive constant of appropriate dimensions. Thena)at points away from the origin, the particle is in unstable equilibriumb)for any finite nonzero value of x, there is a force directed away from the originc)if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)for small displacements from x = 0, the motion is simple harmonicCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.