Physics-
General
Easy

Question

A particle free to move along the x-axis has potential energy given as U open parentheses x close parentheses equals k open square brackets 1 minus exp invisible function application open parentheses negative x to the power of 2 end exponent close parentheses close square brackets blank left parenthesis f o r minus infinity less or equal than blank plus infinity right parenthesis Where k is a positive constant of appropriate dimensions. Then

  1. At points away from origin, the particle is in equilibrium    
  2. For any finite non-zero value of x, there is a force directed away from the origin    
  3. Its total mechanical energy is k divided by 2 and it is equal to its kinetic energy at origin    
  4. At x equals 0, the motion of the particle is simple harmonic    

The correct answer is: At x equals 0, the motion of the particle is simple harmonic


    If a force acting on an object is a function of position only, it is said to be conservative force and it can be represented by a potential energy (U) function which for one dimensional case satisfies the derivative condition.
    F open parentheses x close parentheses equals negative fraction numerator d U over denominator d x end fraction
    Given, U subscript x end subscript equals k open square brackets 1 minus exp invisible function application open parentheses negative x to the power of 2 end exponent close parentheses close square brackets
    therefore F equals negative fraction numerator d U over denominator d x end fraction equals negative 2 k x exp invisible function application left parenthesis negative x to the power of 2 end exponent right parenthesis
    At x equals 0 comma blank F equals 0.
    Hence, at equilibrium force exerted on particle is zero.
    Also, potential energy of the particle is minimum at x equals 0 blank a n d x equals plus-or-minus infinity the potential energy is maximum. Hence, at x equals 0, the motion of particle is simple harmonic.

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