At t - 0- the displacement of a particle in S-H-M- is half its amplitude- Its initial phase is - dfrac-pi -6- rad dfrac-pi -3- rad dfrac-pi -2- rad dfrac-2pi -3- rad

Question

At t - 0- the displacement of a particle in S-H-M- is half its amplitude- Its initial phase is -  dfrac-pi -6- rad dfrac-pi -3- rad dfrac-pi -2- rad dfrac-2pi -3- rad

Toppr Admin · Accepted Answer

Equation of S-H-M-xA0- is y-Asin-x3C9-t-x3D5-at t-0-y-A2-x21D2-12-sin-x3D5-x21D2-x3D5-x3C0-6

At t = 0, the displacement of a particle in S.H.M. is half its amplitude. Its initial phase is :
$\frac{π}{6}$ rad
$\frac{π}{3}$ rad
$\frac{π}{2}$ rad
$\frac{2 π}{3}$ rad

equation of S.H.M is $y = A s i n (ω t + ϕ)$
at $t = 0; y = \frac{A}{2}$
$\Rightarrow \frac{1}{2} = s i n ϕ$
$\Rightarrow ϕ = \frac{π}{6}$

At t = 0, the displacement of a particle in S.H.M. is half its amplitude. Its initial phase is : π6radπ3radπ2rad2π3rad

equation of S.H.M is y=Asin(ωt+ϕ)at t=0;y=A2⇒12=sinϕ⇒ϕ=π6

At t = 0, the displacement of a particle in S.H.M. is half its amplitude. Its initial phase is :
$\frac{π}{6}$ rad
$\frac{π}{3}$ rad
$\frac{π}{2}$ rad
$\frac{2 π}{3}$ rad

equation of S.H.M is $y = A s i n (ω t + ϕ)$
at $t = 0; y = \frac{A}{2}$
$\Rightarrow \frac{1}{2} = s i n ϕ$
$\Rightarrow ϕ = \frac{π}{6}$