The equation of the displacement of two particles making SHM are represented by
$y_{1}$ = a sin $(ω t + ϕ)$ & $y_{2}$ = a cos $(ω t)$ .
The phase difference of the velocities of the two particles is :

$\frac{π}{2} + ϕ$
$- ϕ$
$ϕ - \frac{π}{2}$
$ϕ$

Question

The equation of the displacement of two particles making SHM are represented by
$y_{1}$ = a sin $(ω t + ϕ)$ & $y_{2}$ = a cos $(ω t)$ .
The phase difference of the velocities of the two particles is :

$\frac{π}{2} + ϕ$
$- ϕ$
$ϕ - \frac{π}{2}$
$ϕ$

Toppr Admin · Accepted Answer

Y1-asin-x3C9-t-x3D5-y2-acos-x3C9-t-y1-a-x3C9-cos-x3C9-t-x3D5-a-x3C9-sin-x3C9-t-x3D5-x3C0-2-y2-x2212-a-x3C9-sin-x3C9-t-a-x3C9-sin-x3C0-x3C9-t-therefore phase difference-x3C9-t-x3D5-x3C0-2-x2212-x3C0-x3C9-t-x3D5-x2212-x3C0-2

The equation of the displacement of two particles making SHM are represented by y1 = a sin (ωt+ϕ) & y2 = a cos (ωt). The phase difference of the velocities of the two particles is :π2+ϕ−ϕϕ−π2ϕ

y1=asin(ωt+ϕ)y2=acos(ωt)y1=aωcos(ωt+ϕ)=aωsin(ωt+ϕ+π/2)y2=−aωsin(ωt)=aωsin(π+ωt)therefore phase difference=ωt+ϕ+π/2−(π+ωt)=ϕ−π/2

The equation of the displacement of two particles making SHM are represented by
$y_{1}$ = a sin $(ω t + ϕ)$ & $y_{2}$ = a cos $(ω t)$ .
The phase difference of the velocities of the two particles is :

$\frac{π}{2} + ϕ$
$- ϕ$
$ϕ - \frac{π}{2}$
$ϕ$

$y_{1} = a s i n (ω t + ϕ)$
$y_{2} = a c o s (ω t)$
$y_{1} = a ω c o s (ω t + ϕ) = a ω s i n (ω t + ϕ + π /_{2})$
$y_{2} = - a ω s i n (ω t) = a ω s i n (π + ω t)$
therefore phase difference $= ω t + ϕ + π /_{2} - (π + ω t)$
$= ϕ - π /_{2}$