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 :

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 : (A) (-π/6) (B) (π/3) (C) (-π/3) (D) (π/6)

Tardigrade Admin · Accepted Answer

Given: y1=0.1 sin(100 π t+(π/3)) ∴ (dy1/dt)=v1=0.1×100 π cos(100 π t+(π/3)) or v1=10π sin(100 π t+(π/3)+(π/2)) or v1=10π sin(100 π t+(5π /6)) and y=0.1 cos π t ∴ (dy2/dt)=v2=-0.1 sin π t=0.1 sin (π t+π) Hence, phase difference Δφ=φ1-φ2 =(100 π t+(5π/6))-(π t+π) =(5π/6)-π =(5π /6)-π (at t=0) =-(π/6)

Q. Two simple harmonic motions are represented by the equations $y_{1} = 0.1 s in (100 π t + \frac{π}{3})$ 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 :

$\frac{- π}{6}$

$\frac{π}{3}$

$\frac{- π}{3}$

$\frac{π}{6}$

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 :

6−π​

3π​

3−π​

6π​

Q. Two simple harmonic motions are represented by the equations $y_{1} = 0.1 s in (100 π t + \frac{π}{3})$ 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 :

$\frac{- π}{6}$

$\frac{π}{3}$

$\frac{- π}{3}$

$\frac{π}{6}$