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

During a simple harmonic motion,

Options

  1. Athe average potential energy over an arbitrary time interval equals the average kinetic energy for that same i
  2. Bthe potential energy is never equal to the kinetic energy
  3. Cthe average potential energy over a single time period equals the average kinetic energy for this period
  4. Dthe potential energy is continuously equal to the kinetic energy

Correct answer

C. the average potential energy over a single time period equals the average kinetic energy for this period

Step-by-step solution

In a simple harmonic motion, the displacement and velocity as functions of time are given by x = A ( t) and v = A ( t) . The instantaneous potential energy is U = 1 2 m ^2 x^2 = 1 2 m ^2 A^2 ^2( t) . The instantaneous kinetic energy is K = 1 2 m v^2 = 1 2 m ^2 A^2 ^2( t) . The average potential energy over one complete time period T is: U = 1 T ₀^ T 1 2 m ^2 A^2 ^2( t) dt = 1 4 m ^2 A^2 The average kinetic energy over one complete time period T is: K = 1 T ₀^ T 1 2 m ^2 A^2 ^2( t) dt = 1 4 m ^2 A^2 Thus, the averag

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