The minimum phase difference between two SHM's is:
$y_{1} = sin (\frac{π}{6}) sin (ω t) + sin (\frac{π}{3}) cos (ω t)$
$y_{2} = cos (\frac{π}{6}) sin (ω t) + sin (\frac{π}{3}) cos (ω t)$ is:
$\frac{π}{3}$
$\frac{π}{6}$
$\frac{π}{12}$
$0$

Question

The minimum phase difference between two SHM's is:
$y_{1} = sin (\frac{π}{6}) sin (ω t) + sin (\frac{π}{3}) cos (ω t)$
$y_{2} = cos (\frac{π}{6}) sin (ω t) + sin (\frac{π}{3}) cos (ω t)$ is:
$\frac{π}{3}$
$\frac{π}{6}$
$\frac{π}{12}$
$0$

Toppr Admin · Accepted Answer

Y1-sin-x3C9-t-sin-x3C0-6-cos -x3C9-tsin-x3C0-3-sin -x3C9-tcos-x3C0-3-cos -x3C9-tsin-x3C0-3-sin-x3C9-t-x3C0-3-Similarly for the second equation is-y2-sin-x3C9-t-x3C0-6-

The minimum phase difference between two SHM's is:y1=sin(π6)sin(ωt)+sin(π3)cos(ωt)y2=cos(π6)sin(ωt)+sin(π3)cos(ωt) is:π3π6π120

y1=sin(ωt)sin(π6)+cos ωtsinπ3=sin ωtcos(π3)+cos ωtsinπ3=sin(ωt+π3)Similarly for the second equation is:y2=sin(ωt+π6)

The minimum phase difference between two SHM's is:
$y_{1} = sin (\frac{π}{6}) sin (ω t) + sin (\frac{π}{3}) cos (ω t)$
$y_{2} = cos (\frac{π}{6}) sin (ω t) + sin (\frac{π}{3}) cos (ω t)$ is:
$\frac{π}{3}$
$\frac{π}{6}$
$\frac{π}{12}$
$0$

$y_{1} = sin (ω t) sin (\frac{π}{6}) + cos$ $ω t sin \frac{π}{3}$

$= sin$ $ω t cos (\frac{π}{3}) + cos$ $ω t sin \frac{π}{3}$

$= sin (ω t + \frac{π}{3})$

Similarly for the second equation is:
$y_{2} = sin (ω t + \frac{π}{6})$