Two SHM is represented by y1=0.1 sin [100πt+π3  ]  and y2=0.1 cos πt. The phase difference of the velocity of particle 1 with respect to 2 is

The phase difference between two simple harmonic motions (SHMs) can be found by comparing their equations.

The given equations are:

y1 = 0.1 sin (100πt + π/3) and y2 = 0.1 cos πt

First, we need to find the velocities of the two particles. The velocity of a particle in SHM is given by the derivative of its displacement equation with respect to time.

For particle 1, the velocity v1 is the derivative of y1 with respect to time:

v1 = dy1/dt = 0.1 * 100π * cos(100πt + π/3)

For particle 2, the velocity v2 is the derivative of y2 with respect to time:

v2 = dy2/dt = -0.1 * π * sin(πt)

The phase difference between the two velocities can be found by comparing the arguments of the cosine and sine functions in the velocity equations.

The argument of the cosine function in v1 is (100πt + π/3), and the argument of the sine function in v2 is πt.

Therefore, the phase difference is (100πt + π/3) - πt = 99πt + π/3.

This is the phase difference of the velocity of particle 1 with respect to particle 2.