The displacement vs time of a particle executing SHM is shown in figure. The initial phase φ is

Question

The displacement vs time of a particle executing SHM is shown in figure. The initial phase φ is (A) -π<φ<-(π/2) (B) π<φ<(3 π/2) (C) -(3 π/2)<φ<-π (D) (π/2)<φ<π

Tardigrade Admin · Accepted Answer

For x=(- A ), we have -A=A sin (ω × 0+φ0) or φ0=-(π/2). So for x<(-A), φ0<(-π / 2).

Q. The displacement vs time of a particle executing SHM is shown in figure. The initial phase $ϕ$ is

$- π < ϕ < - \frac{π}{2}$

$π < ϕ < \frac{3 π}{2}$

$- \frac{3 π}{2} < ϕ < - π$

$\frac{π}{2} < ϕ < π$

For $x = (- A)$ ,
we have $- A = A sin (ω \times 0 + ϕ_{0})$
or $ϕ_{0} = - \frac{π}{2}$ .
So for $x < (- A), ϕ_{0} < (- π /2)$ .

Q. The displacement vs time of a particle executing SHM is shown in figure. The initial phase ϕ is

−π<ϕ<−2π​

π<ϕ<23π​

−23π​<ϕ<−π

2π​<ϕ<π