Home
Class 12
PHYSICS
The potential energy of a particle of ma...

The potential energy of a particle of mass `0.02` `kg` moving along `x`-axis is given by `V` `=` `Ax(x - 4)` `J` where `x` is in metres and `A` is a constant. Which of the following is/are correct statement (s)?

`(A)` The particle is acted upon by a constant force.
`(B)` The particle executes simple harmonic motion.
`(C)` The speed of the particle is maximum at `x = 2` `m.`
`(D)` The period of oscillation of the particle is `pi/5` sec.

A

The particle is acted upon by a constant force.

B

The particle executes simple harmonic motion.

C

The speed of the particle is maximum at x = 2 m.

D

The period of oscillation of the particle is `pi/5` sec.

Text Solution

Verified by Experts

Promotional Banner

Topper's Solved these Questions

  • QUESTION PAPER 2021

    WB JEE PREVIOUS YEAR PAPER|Exercise PHYSICS (CATEGORY -II)|6 Videos

  • QUESTION PAPER 2019

    WB JEE PREVIOUS YEAR PAPER|Exercise PHYSICS (CATEGORY -III)|3 Videos

Similar Questions

Explore conceptually related problems

Show that the total mechanical energy remains constant for a particle executing simple harmonic motion.

Write the equation of motion of a particle executing simple harmonic motion, whose displacement is maximum at t = 0.

Find expressions for potential and kinetic energies of a particle executing simple harmonic motion.

Derive an expression for the total energy of a particle undergoing simple harmonic motion.

A moving particle has an acceleration a = -bx, where b is a positive constant and x is the distance of the particle from the equilibrium position. What is the time period of oscillation of the particle?

A particle executives simple harmonic motion of amplitude A. The distance from the mean position where its kinetic energy is equal to its potential energy is