A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy – time graph of the particle will look like:

In an engine the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and I, respectively. At time t = 0 one particle has displacement A while the other one has displacement $${{ - A} \over 2}$$ and they are moving towards each other. If they cross each other at time t, then t is :

A particle performs simple harmonic motion with amplitude $$A.$$ Its speed is trebled at the instant that it is at a distance $${{2A} \over 3}$$ from equilibrium position. The new amplitude of the motion is: