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
$x = 2 sin 4 π t$
$x = 2 sin 2 π t$
$x = sin 2 π t$
$x = 4 sin 2 π t$

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
$x = 2 sin 4 π t$
$x = 2 sin 2 π t$
$x = sin 2 π t$
$x = 4 sin 2 π t$

Toppr Admin · Accepted Answer

Max- displacement - Amplitude - 2cm- thus it-apos-s of the form x-2sin-x3C9-t -assuming initial phase as 0-Now- -x3C9-2-x3C0-T-2-x3C0-1-2-x3C0-Thus- x-2sin2-x3C0-t

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 bex=2sin4πtx=2sin2πtx=sin2πtx=4sin2πt

Max. displacement = Amplitude = 2cm, thus it's of the form x=2sinωt (assuming initial phase as 0)Now, ω=2π/T=2π/1=2πThus, x=2sin2πt

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
$x = 2 sin 4 π t$
$x = 2 sin 2 π t$
$x = sin 2 π t$
$x = 4 sin 2 π t$

Max. displacement = Amplitude = 2cm, thus it's of the form $x = 2 s i n ω t$ (assuming initial phase as 0)
Now, $ω = 2 π / T = 2 π / 1 = 2 π$
Thus, $x = 2 s i n 2 π t$