Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Manipal MET2012PhysicsOscillations

For a particle executing simple harmonic motion, the kinetic energy K is given by, K=K₀ ^2 t . The maximum value of potential energy is

Options

  1. AK₀
  2. Bzero
  3. CK₀ / 2
  4. Dnot obtainable

Correct answer

A. K₀

Step-by-step solution

K_ =K₀= total energy As total energy remains conserved in SHM, hence when U is maximum in SHM , K=0 , i.e., E iş also equal to U_ , i.e., U_ =E=K₀ .

Practice Oscillations on Quantrex Academy →

More from Oscillations

Force constant of interatomic bond, in a certain element, is 7.1 Nm⁻¹ . If the atom oscillates in SHM in a certain direction, what is its frequency? Given: Mole weight of the given 2026An object of mass 10g executes SHM along the x -axis with frequency of ( 10 ) Hz. At the point x = 2 cm the object has KE 1.2 J and PE 0.4 J. The amplitude of oscillation is 2026A simple pendulum of length L , mass m and electric charge q on its bob is oscillating with a time period T under uniform gravity which is in the - z direction. Upon applying a uni 2026A sphere and a cube of equal masses on a horizontal frictionless floor, are confined between two vertical walls, as shown in the figure. The cube is attached to the wall by a massl 2026A mass of 1 kg is executing SHM. Its displacement is given by x = 6.0 (100t + /4) cm. What is the maximum kinetic energy? 2026If the displacement of a particle executing simple harmonic motion is given by x =0.5 (125.6 t ) , then the time period of oscillation of the particle is nearly (Here x is displace 2025The amplitude of a damped harmonic oscilator becomes 50 % of its initial value in a time of 12 s. If the amplitude of the oscillator at a time of 36 s is x % of its initial amplitu 2025If the function ^2 t ( t is time in second) represents a periodic motion, then the period of the motion is 2025 Full Oscillations list All Manipal MET PYQs