Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JIPMER2018PhysicsOscillations

A mass of Hg is executing SHM which is given by x=6.0 (100 t+ 4 ) cm . What is the maximum kinetic energy?

Options

  1. A3 J
  2. B6 J
  3. C9 J
  4. D18 J

Correct answer

D. 18 J

Step-by-step solution

Here, m =1 ~kg The given equation of SHM is x=6.0 (100 t+ 4 ) Comparing it with equation of SHM, aligned x & =A ( t+ ), we have A & =6.0 ~cm = 6 100 ~m aligned and =100 rad / s Maximum kinetic energy = 1 2 ~m ( v _ )^2= 1 2 ~m ( ~A )^2= 1 2 1 [ 6 100 100 ]^2=18 ~J

Practice Oscillations on Quantrex Academy →

More from Oscillations

A simple pendulum oscillating in air has a period of 3 second. If it is immersed in a non-viscous liquid, having density = ( 1 x ) , where is the density of the material of the bob 2026Identify the equation that DOES NOT represent oscillatory motion. ( A is amplitude of motion) 2026The potential energy of a simple harmonic oscillator when the particle is at 3 4 A , is ( A is the amplitude of oscillation and E is the total energy of the oscillator) 2026Two SHM's x₁ = a ( t + 3 2 ) and x₂ = a ( t + ) are superimposed on each other. The resultant amplitude of motion is ( /2 = 1 and /2 = 0) 2026The average acceleration of a particle performing S.H.M. over two complete oscillations is (A = amplitude, = angular frequency) 2026The periodic time of a simple pendulum, (the simple pendulum with its bob is in air) is T. But if the bob of the simple pendulum is completely immersed in a non viscous liquid, who 2026The graph shows variation of displacement of a particle performing SHM with time t. which of the following statements is correct from the graph? 2026Two particles A and B execute S.H.M.s of periods T and 5T 2 . If they start from the mean position, then the phase difference between them when the particle A completes two oscilla 2026 Full Oscillations list All JIPMER PYQs