Given, Amplitude of a SHM = a Time period of SHM =T The displacement of a particle in case of SHM is given by, x=a‌sin(ωt+ϕ) then the velocity of SHM is given as ∴v=
dx
dt
or v=
d
dt
[a‌sin(ωt+ϕ)] or v=aω[1−sin2(ωt+ϕ)]1∕2 or =aω[1−
x2
a2
]1∕2 or v=ω√a2−x2 Now, at mean position (x=0) velocity is maximum. i.e. vmax=aω ........(i) We know that, angular frequency oscillating particle is ω=
2Ï€
T
.........(ii) Now, putting the value of omega is Eq. (i), we get vmax=a×