A particle executes simple harmonic motion with frequency 2.5 Hz and amplitude 2 m. The speed of the particle 0.3 s after crossing the equilibrium position is:

Two particles executing SHM of the same amplitude and frequency on parallel lines side by side. They cross one another when moving in opposite directions each time their displacement is $\dfrac{\sqrt{3}}{2}$ times their amplitude. The phase difference between them is:

একজন বালিকা একটি দোলনায় বসে দোল খাচ্ছে। বালিকাটি উঠে দাঁড়ালে দোলনকালের কি পরিবর্তন হবে? 

The time period of a simple pendulum is air is $T$. Now the pendulum is submerged in a liquid of density $\dfrac{\rho}{16}$ where $\rho$ is density of the bob of the pendulum. The new time period of oscillation is.

The amplitude and the periodic time of a SHM are $5cm$ and $6s$ respectively. At a distance of $2.5cm$ away from the mean position, the phase will be