Two simple harmonic motions are represented by the equations y1=0.1 sin (100 π t+(π/3)) and y2=0.1 cos π t . The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is at t=0

Question

Two simple harmonic motions are represented by the equations y1=0.1 sin (100 π t+(π/3)) and y2=0.1 cos π t . The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is at t=0 (A) (-π/3) (B) (π/6) (C) (-π/6) (D) (π/3)

Tardigrade Admin · Accepted Answer

y1 =0.1 sin (100 π t+(π/3)) v1 =(d y1/d t)=0.1 × 100 π cos (100 π t+(π/3)) =10 π cos (100 π t+(π/3)) =10 π sin ((π/2)+100 π t+(π/3)) y2 =0.1 cos π t v2 =(d y2/d t)=-0.1 π sin π t =0.1 π sin (π+π t) At t =0, phase difference of v1 w.r.t. v2 =((π/2)+(π/3))-π=-(π/6)

Q. Two simple harmonic motions are represented by the equations $y_{1} = 0.1 sin (100 π t + \frac{π}{3})$ and $y_{2} = 0.1 cos π t .$ The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is at $t = 0$

$\frac{- π}{3}$

$\frac{π}{6}$

$\frac{- π}{6}$

$\frac{π}{3}$

Q. Two simple harmonic motions are represented by the equations y1​=0.1sin(100πt+3π​) and y2​=0.1cosπt. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is at t=0

3−π​

6π​

6−π​

3π​

y1​=0.1sin(100πt+3π​) v1​=dtdy1​​=0.1×100πcos(100πt+3π​) =10πcos(100πt+3π​) =10πsin(2π​+100πt+3π​) y2​=0.1cosπt v2​=dtdy2​​=−0.1πsinπt =0.1πsin(π+πt) At t=0, phase difference of v1​ w.r.t. v2​ =(2π​+3π​)−π=−6π​

Q. Two simple harmonic motions are represented by the equations $y_{1} = 0.1 sin (100 π t + \frac{π}{3})$ and $y_{2} = 0.1 cos π t .$ The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is at $t = 0$

$\frac{- π}{3}$

$\frac{π}{6}$

$\frac{- π}{6}$

$\frac{π}{3}$