A particle is executing S.H.M. with an amplitude of 4 cm and time period 12 seconds.  The time taken by the particle in going 2 cm from the mean position is T₁ seconds.  The time taken from this displaced position to reach the extreme position on the same side is T₂ seconds. Then T₁ /T₂ will be

A particle starts simple harmonic motion from the mean position.  Its amplitude is a and time period is T.  What is its displacement when its speed is half of its maximum speed ?

A particle is subjected to two mutually perpendicular simple harmonic motions such that is x and y coordinates are given by


The path of the particle will be

A horizontal surface moves up and down in S.H.M with an amplitude of 1 cm.  If mass of 10 kg (which is placed on the surface) is to remain continuously in contact with it, the maximum frequency of S.H.M. will be

Two springs of constants k₁ and k₂ have equal maximum velocities when executing simple harmonic motion.  The ratio of their amplitudes (masses are equal) will be

A particle starts S.H.M from the mean position.  Its amplitude is A and time period is T.  At the time its speed is half of the maximum speed.  Its displacement y from the mean position is

A mass m is suspended from a spring of negligible mass and the system oscillates with a frequency n.  What will be the frequency of oscillations if a mass of 4 m is suspended from the same spring ?

The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position.  If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of oscillations is