Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Concepts Of Physics MCQ Edition [Volume 1]PhysicsSimple Harmonic Motion

A student claims that he applied a force F = -k x on a particle, resulting in the particle executing simple harmonic motion. He declines to reveal whether k is a constant or not. Assume that he has worked exclusively with positive x and that no other force acted on the particle.

Options

  1. AThe motion cannot be simple harmonic.
  2. BAs x increases, k remains constant.
  3. CAs x increases, k decreases.
  4. DAs x increases, k increases.

Correct answer

D. As x increases, k increases.

Step-by-step solution

For a particle to execute simple harmonic motion, the restoring force must be of the form F = -c(x - x₀) , where c > 0 is a constant and x₀ is the equilibrium position. Given that the applied force is F = -k x , we can equate the two expressions: -k x = -c(x - x₀) k = c(x - x₀) x = c ( x - x₀ x ) To determine the variation of k with respect to x , we differentiate k with respect to x : dk dx = c 2 x + cx₀ 2x x Since the particle executes SHM exclusively in the region x > 0 , its equilibrium position x₀ must also be

Practice Simple Harmonic Motion on Quantrex Academy →

More from Simple Harmonic Motion

At a specific instant, the magnitudes of the position, velocity, and acceleration of a particle undergoing simple harmonic motion are observed to be 2 cm , 1 m/s , and 10 m/s ^2 , For a particle executing simple harmonic motion, the maximum speed and acceleration are 10 cm/s and 50 cm/s ^2 , respectively. Determine the position(s) of the particle when its spA particle having a mass of 10 g oscillates according to the equation x = (2.0 cm ) [(100 s ⁻¹) t + /6] . Determine the amplitude, the time period, and the spring constant for thisA particle undergoes simple harmonic motion having an amplitude of 10 cm . At what distance from the mean position will its kinetic and potential energies be equal?A particle executes simple harmonic motion with an amplitude of 10 cm and a time period of 6 s . At t = 0 , it is located at x = 5 cm and is moving towards the positive x-directionA particle having a mass of 10 g oscillates according to the equation x = (2.0 cm ) [(100 s ⁻¹) t + /6] . Determine the position, the velocity, and the acceleration of the particleA particle begins its motion at t = 0 according to the equation x = 5 (20 t + /3) , where x is in centimetre and t is in second. At what time will the particle first have zero acceA particle starts moving at t = 0 with its position given by the equation x = 5 (20 t + /3) , where x is in centimetre and t is in second. Determine the time when the particle firs Full Simple Harmonic Motion list All Concepts Of Physics MCQ Edition [Volume 1] PYQs