Two simple harmonic motions are represented by the equations y1-0-1 sin 100 - t-3 and y2-0-1 cos- t- The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

V_1=dy_1dt=0.1 × 100 πcos ( 100 π t +π3 ) v_2=dy_2dt= 0.1 πsinπ t=0.1 πcos ( π t+π2 ) The phase difference of velocity of the first particle with respect to the ...

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Standard XIII

Physics

Phase Difference

Two simple ha...

Question

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

\frac{- π}{3}

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\frac{π}{6}

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\frac{- π}{6}

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\frac{π}{3}

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Solution

The correct option is C $\frac{- π}{6}$
$v_{1} = \frac{d y_{1}}{d t} = 0.1 \times 100 π cos (100 π t + \frac{π}{3})$
$v_{2} = \frac{d y_{2}}{d t} = - 0.1 π sin π t = 0.1 π cos (π t + \frac{π}{2})$
The phase difference of velocity of the first particle with respect to the velocity of 2nd particle at $t = 0$ is
$Δ φ = φ_{1} - φ_{2} = \frac{π}{3} - \frac{π}{2} = - \frac{π}{6}$

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