Two simple harmonic motions are represented by the equations Y1 - 0-1 sin100 - t - -3 and Y2 - 0-1 cos - t- The phase of velocity of particle 1 with respect to the velocity of particle 2 at time t - 0 is

Velocity of the particles can be calculated as v_1 = dy_1dt = 0.1 × 100 π cos(100 π t + π3) v_2 = dy_2dt = 0.1 π sin π t=0.1π cos(π t + π2) Phase difference o ...

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Standard XII

Physics

Phase Difference

Two simple ha...

Question

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

\frac{- 2 π}{3}

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\frac{π}{3}

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\frac{- π}{6}

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\frac{2 π}{3}

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Solution

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

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Two simple harmonic motions are represented by the equations Y1=0.1 sin(100 π t+π3) and Y2=0.1 cos π t. The phase of velocity of particle 1 with respect to the velocity of particle 2 at time t=0 is

The correct option is C −π6Velocity of the particles can be calculated as v1=dy1dt=0.1×100 π cos(100 π t+π3) v2=dy2dt=−0.1 π sin π t=0.1π cos(π t+π2) Phase difference of velocity of first particle with respect to the velocity of particle 2 at t=0 is Δϕ=ϕ1−ϕ2=π3−π2=−π6.

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