Two simple harmonic motions are represented by the equations y1=0.1 sin (100 π t+(π/3)) and y1=0.1 cos π t The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is - examsnet.com

Question

Two simple harmonic motions are represented by the equations y1=0.1 sin (100 π t+(π/3)) and y1=0.1 cos π t The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is - examsnet.com (A) (π/3) (B) (-π/6) (C) (π/6) (D) (-π/3)

Tardigrade Admin · Accepted Answer

v1=(d y1/d t)=0.1 × 100 π cos (100 π t+(π/3)) v2=(d y2/d t)=-0.1 π sin π t=0.1 π cos (π t+(π/2)) ∴ Phase diff. = varphi1- varphi2 =(π/3)-(π/2)=(2 π-3 π/6)=-(π/6)

Q. Two simple harmonic motions are represented by the equations $y_{1} = 0.1 sin (100 π t + \frac{π}{3})$ and $y_{1} = 0.1 cos π t$ The phase difference of the velocity of particle 1 with respect to the velocity of particle $2$ is - examsnet.com

$\frac{π}{3}$

$\frac{- π}{6}$

$\frac{π}{6}$

$\frac{- π}{3}$

$v_{1} = \frac{d y _{1}}{d t} = 0.1 \times 100 π cos (100 π t + \frac{π}{3})$
$v_{2} = \frac{d y _{2}}{d t} = - 0.1 π sin π t = 0.1 π cos (π t + \frac{π}{2})$
$∴$ Phase diff. $= φ_{1} - φ_{2}$
$= \frac{π}{3} - \frac{π}{2} = \frac{2 π - 3 π}{6} = - \frac{π}{6}$

Q. Two simple harmonic motions are represented by the equations y1​=0.1sin(100πt+3π​) and y1​=0.1cosπt The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is - examsnet.com

3π​

6−π​

6π​

3−π​