A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $${1 \over 4}$$ times that of the bob and the length of the thread is increased by 1/3^rd of the original length, then the time period of the simple harmonic oscillations will be :-

For a body executing S.H.M. :

(1) Potential energy is always equal to its K.E.

(2) Average potential and kinetic energy over any given time interval are always equal.

(3) Sum of the kinetic and potential energy at any point of time is constant.

(4) Average K.E. in one time period is equal to average potential energy in one time period.

Choose the most appropriate option from the options given below :

The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure.


The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure :

An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $$t = {T \over 4}s$$ starting from mean position. Assume that the initial phase of the oscillation is zero.