Two simple harmonic motions are represented by the equations y 1-0-1 sin100 - t -3 and y 2-0-1 cost tThe phase difference of the velocity of particle 1 with respect to the velocity of particle 2 at t -0 is A- 3-3B- 2 -6 -C- -3D- 1-6

The correct option is B v_1=dy_1/dt=0.1× 100 π cos(100π t+π/3) v_2=dy_2/dt= 0.1 π sin π t=0.1π cos(π t+π/2) Phase difference of velocity of first particle with ...

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Standard XII

Physics

Phase Difference

Two simple ha...

Question

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

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Solution

The correct option is C
$v_{1} = \frac{d y_{1}}{d t} = 0.1 \times 100 π c o s (100 π t + \frac{π}{3})$
$v_{2} = \frac{d y_{2}}{d t} = - 0.1 π s i n π t = 0.1 π c o s (π t + \frac{π}{2})$
Phase difference of velocity of first particle with respect to the velocity of $2^{n d}$ particle at
t = 0 is
$△ ϕ = ϕ_{1} - ϕ_{2} = \frac{π}{3} - \frac{π}{2} = - \frac{π}{6} .$

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Two simple harmonic motions are represented by the equations y1= 0.1 sin(100π t+π3) and y2 = 0.1 cosπ tThe phase difference of the velocity of particle 1 with respect to the velocity of particle 2 at t= 0 is

The correct option is C v1=dy1dt=0.1× 100 π cos(100π 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 2nd particle at t = 0 is △ ϕ=ϕ1−ϕ2=π3−π2=−π6.

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