Question

The maximum displacement of a particle executing SHM from its mean position is 2 cm and its time period is 1 s. The equation of its displacement will be

• A. $x=2\sin 4\pi t$
• B. $x=2\sin 2\pi t$
• C. $x=\sin 2\pi t$
• D. $x=4\sin 2\pi t$

Answer 1

The correct option is B $x=2\sin 2\pi t$

Max. displacement = Amplitude = 2cm, thus it's of the form $x=2sin\omega t$ (assuming initial phase as 0)
Now, $\omega =2\pi /T=2\pi /1=2\pi$
Thus, $x=2sin2\pi t$