Time period calculation in SHM

The figure shows the displacement-time graph of a particle executing SHM . If the time period of oscillation is 2s , then the equation of motion is given by

Time period of a particle in SHM depends on the force constant k and mass m of the particle T = 2psqrt((m)/(k)) . A simple pendulum exeutes SHM approximately. Why then is the period of a pendulum independent of the mass of the pendulum ?

The acceleration versus displacement graph of a particle performing SHM is shown in figure. The time period of oscillation of particle in second is

A body of mass 10g executes SHM with amplitude 2 xx 10^(-2) m and time period 2 s Calculate the energy of the particle.

A particle is in SHM with time period 3 sec. Calculate the displacement of particle from mean position after t = 0.5 sec.

If displacement x and velocity v related as 4v^(2) = 25 - x^(2)m in a SHM Then time period of given SHM is (consider SI unit)

The time period of an oscillating body executing SHM is 0.05 sec and its amplitude is 40 cm. The maximum velocity of particle is :